Applications of Forces

Transcription

1 Chapter 10 Applications of orces Practice Problem Solutions Student Textbook page (a) rame the Problem - Make a sketch of the ector. - The angle is between 0 and 90 so it is in the first quadrant. Therefore, both the x-component and the y-component will be positie. - Use trigonometric functions to find components of the ector. The components d x, d y of ector d Variables and Constants Known Unknown d 16 m d x θ 75 d y Strategy Draw the ector with its tail at the origin of an x-y-coordinate system. Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. Determine the signs of the components d x d cos θ d x 16 m cos 75 d x 4.14 m d y d sin θ d y 16 m sin 75 d y m The x-component lies on the positie x-axis so it is positie. The y-component lies on the positie y-axis so it is positie. (a) The x-component of the ector is +4.1 m and the y-component of the ector is +15 m. Validate Use the Pythagorean theorem to check your answers. d d x + d y d (4.14 m) + (15.45 m) d d 16 m The alue agrees with original ector. 4

2 1. (b) rame the Problem - Make a sketch of the ector. - The angle is between 90 and 180 so it is in the second quadrant. Therefore, the x-component will be negatie and the y-component will be positie. - Use trigonometric functions to find components of the ector. The components a x, a y of ector a. Variables and Constants Known Unknown a 8.1 m a s x θ 145 a y Strategy Draw the ector with its tail at the origin of an x-y-coordinate system. Identify the angle with the closest x-axis and label it θ R Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. θ R θ R 35 a x a cos θ R a x 8.1 m cos 35 s Determine the signs of the components a x m s a y a sin θ R a y 8.1 m sin 35 s a y m s The x-component lies on the negatie x-axis so it is negatie. The y-component lies on the positie y-axis so it is positie. (b) The x-component of the ector is 6.6 m/s and the y-component of the ector is +4.6 m/s. Validate Use the Pythagorean theorem to check your answers. a ax + a y a (6.635 m ) + (4.646 m ) s s a ( m ) s a 8.1 m s The alue agrees with the magnitude of the original ector. 1. (c) rame the Problem - Make a sketch of the ector. - The angle is between 180 and 70 so it is in the third quadrant. Therefore, the x-component will be negatie and the y-component will be negatie. - Use trigonometric functions to find components of the ector. The components x, y of ector. 43

3 Variables and Constants Known Unknown 16.0 m s x θ 5 y Strategy Draw the ector with its tail at the origin of an x-y-coordinate system. Identify the angle with the closest x-axis and label it θ R Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. θ R θ R 45 x cos θ R x 16.0 m cos 45 s x m s y sin θ R y 16.0 m sin 45 s y m s Determine the signs of the components The x-component lies on the negatie x-axis so it is negatie. The y-component lies on the negatie y-axis so it is negatie. (c) The x-component of the ector is 11.3 m/s and the y-component of the ector is 11.3 m/s. Validate Use the Pythagorean theorem to check your answers. x + y ( m s ) + ( m s ) ( m s ) 16.0 m s The alue agrees with the magnitude of the original ector.. (a) rame the Problem - Make a sketch of the ector. - The ector is described in compass directions. - Conert to an x-y-coordinate system. Let +y be north, and y south. East will become +x and west x. - You will need to find the angle with the closest x-axis. The components d x, d y of ector d 0.0 km[n0.0 E]. Variables and Constants Known d 0.0 km[n0.0 E] Unknown d x d y θ 44

4 Strategy Draw an x-y-coordinate system and indicate that the axes also represent compass directions. Draw the ector with its tail at the origin. Identify the angle, θ. θ θ 70 Use the 0.0 made by the ector to the y-axis to find the angle the ector makes with the x-axis. Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. d x d cos θ d x 0.0 km cos 70 d x 6.84 km Determine the signs of the components d y d sin θ d y 0.0 km sin 70 d y km The x-component lies on the positie x-axis so it is positie. The y-component lies on the positie y-axis so it is positie. (a) The x-component of the ector is km and the y-component of the ector is km. Validate Use the Pythagorean theorem to check your answers. d d x + d y d (6.840 km) + ( km) d d 0.0 km The alue agrees with original ector.. (b) rame the Problem - Make a sketch of the ector. - The ector is described in compass directions. - Conert to an x-y-coordinate system. Let +y be north, and y south. East will become +x and west will be x. - You will need to find the angle with the closest x-axis. The components x, y of ector 3.0 m/s[e30.0 S] Variables and Constants Known 16.0 m s [E30.0 S] Unknown x y θ 45

5 Strategy Draw an x-y-coordinate system and indicate that the axes also represent compass directions. Draw the ector with its tail at the origin. Identify the angle, θ. The ector makes an angle θ 30.0 with the x-axis. Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. x cos θ x 3.0 m cos 30 s x.598 m s y sin θ y 3.0 m sin 30 s y 1.50 m s Determine the signs of the components The x-component lies on the positie x-axis so it is positie. The y-component lies on the negatie y-axis so it is negatie. (b) The x-component of the ector is.6 m/s and the y-component of the ector is 1.5 m/s. Validate Use the Pythagorean theorem to check your answers. x + y (.598 m s ) + (1.50 m s ) ( m s ) 3.0 m s The alue agrees with the magnitude of the original ector.. (c) rame the Problem - Make a sketch of the ector. - The ector is described in compass directions. - Conert to an x-y-coordinate system. Let +y be north, and y south. East will become +x and west will be x. - You will need to find the angle with the closest x-axis. The components x, y of ector 6.8 m/s[w70.0 N] Variables and Constants Known 6.8 m s [W70.0 N] Unknown x y θ 46

6 Strategy Draw an x-y-coordinate system and indicate that the axes also represent compass directions. Draw the ector with its tail at the origin. Identify the angle, θ. The ector makes an angle θ 70.0 with the x-axis. Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. x cos θ x 6.8 m cos 70 s Determine the signs of the components x.36 m s y sin θ y 6.8 m sin 70 s y m s The x-component lies on the negatie x-axis so it is negatie. The y-component lies on the positie y-axis so it is positie. (c) The x-component of the ector is.3 m/s and the y-component of the ector is +6.4 m/s. Validate Use the Pythagorean theorem to check your answers. x + y (.36 m s ) + (6.390 m s ) 46.4( m s ) 6.8 m s The alue agrees with the magnitude of the original ector. 3. rame the Problem - Make a sketch of the situation. - The position of the balloon is described in compass directions. - Resoling the ector into components will allow you to determine how the pickup ehicle should reach the balloon. - Conert to an x-y-coordinate system. Let +y be north, and y south. East will become +x and west will be x. - You will need to find the angle with the closest x-axis. The components d x, d y of ector d 60.0 km[e60.0 N]. Variables and Constants Known d 60.0 km[e60.0 N] Unknown d x d y θ 47

7 Strategy Draw an x-y-coordinate system and indicate that the axes also represent compass directions. Draw the ector with its tail at the origin. Identify the angle, θ. The ector makes an angle of θ 60.0 with the positie x-axis. Draw lines from the tip of the ector to each axis, so that they are parallel to the axes. Calculate the components. d x d cos θ d x 60.0 km cos 60 d x 30.0 km d y d sin θ d y 60.0 km sin 60 d y km Determine the signs of the components The x-component lies on the positie x-axis so it is positie. The y-component lies on the positie y-axis so it is positie. The pickup ehicle will hae to drie km east and then 5 km north to pick up the balloon. Validate Use the Pythagorean theorem to check your answers. d dx + d y d (30.0 km) + (51.96 km) d d 60.0 km Practice Problem Solutions Student Textbook page rame the Problem - Make a diagram of the problem. - The hiker s trip is in three segments which are gien in terms of displacements and compass directions. You can conert the compass directions to the x-y-coordinate system. - The total displacement is the ector sum of the three displacement ectors. - The components of each displacement can be determined and summed to get the components of the resultant ector. (a) The displacement of the boat, d total (b) The direction, θ, the boat would need in order to head straight home 48

8 Variables and Constants Known d A.7 km[s] Unknown d total d B 3.4 km[s6 E] d Ax d Ay d C 1.9 km[e1 N] d Bx d By d Cx d Cy θ θ A θ B θ C Strategy Draw the first displacement on an x-y-coordinate system with +y corresponding to north. ind the angle the ector makes Because displacement A lies along the with the x-axis. negatie y-axis, the angle θ A 0. Calculate the x- and y-components Thus, the x-component will be zero, and of displacement A and determine y-component will be negatie. the signs of the components. Draw displacement B and calculate d Ax 0 its x- and y-components. d Ay.7 km Determine the angle the ector θ B 90 6 makes with the x-axis, θ B. θ B 64 Calculate the x- and y-components d Bx d B cos θ of displacement B. d Bx 3.4 km cos 64 d Bx km Determine the signs of the components. Draw displacement C and calculate its x- and y-components. Determine the angle the ector makes with the x-axis, θ C. Calculate the x- and y-components of displacement B. Determine the signs of the components. The components of the resultant ector can be summed indiidually. d By d B sin θ d By 3.4 km sin 64 d By km The ector lies in the 4th quadrant, so the x-component is positie and the y-component is negatie. θ C 1 d Cx d C cos θ d Cx 1.9 km cos 1 d Cx km d Cy d C sin θ d Cy 1.9 km sin 1 d Cy km The ector lies in the first quadrant, so the x- and y-components are positie. d total x d Ax + d Bx + d Cx d total x km km d total x km d total y d Ay + d By + d Cy d total y km km d total y km 49

9 Use the Pythagorean theorem to ( d total ) ( d total x ) + ( d total y ) determine the magnitude of the ( d total ) (3.348 km) + ( km) resultant displacement. ( d total ) km d total 6.30 km Use the components to determine the direction of the resultant displacement. (a) The displacement is 6.3 km[e58 S]. tan θ km km tan θ θ tan θ (b) To head straight back, the boat should head in the direction [W58 N] Validate In each case, the units cancelled to gie the correct units for the desired quantity. The displacement, 6.3 km, is less than the total distance traelled (.7 km km km), as it should be. The solution can be checked using a graphical method. 5. rame the Problem -Make a diagram of the problem. - The jet-ski wants to trael to an island and must aim further uprier to compensate for the current which is flowing to the east. - The elocity of the jet-ski relatie to the shore, js, is the ector sum of the elocity of the rier relatie to the shore, rs, and the elocity of the jet-ski relatie to the rier, jr. The magnitude of the elocity of the jet-ski relatie to the rier, jr, is unknown, but its direction is known. -In other words, the resultant elocity is the sum of the jet ski s heading, and the rier elocity. The magnitude of the jet-ski s heading is gien, but not its direction. Conersely, the direction of the resultant elocity is gien, but not its magnitude. - The sine law can be used. (a) The direction the jet-ski must head in order to trael to the island. (b) The time it will take him to reach the island, t. Variables and Constants Identify the Variables Known Implied Unknown jr km / h[ θ] θ + ϕ 0 θ ϕ rs 60. km / h[e] d 50. km / h[w0.0 S] js t θ 0 jr 40.0 km/h ϕ js 160 θ 0 rs 6.0 km/h[e] 50

10 Deelop a Strategy Determine how the ectors add together. Note that it would be difficult to use components here because we know the magnitude of jr but not its direction (θ) and we know the direction of js but not its magnitude. + js jr rs rom the figure, it is eident that the sine law can be used to sole for the unknown elocity, js, and the unknown direction, θ. Note that θ + ϕ 0 Sole for θ. sinϕ sin160 6 km / h 40 km / h 6 km / h sinϕ sin km / h sin ϕ ϕ sin ( ) ϕ 94. θ 0 ϕ θ θ (a) He should head the jet-ski in the θ direction [W17.1 S]. Apply the sine law again to find the sinθ sin160 unknown elocity, js 40 km / h js sinθ js ( 40 km / h) sin160 sin js ( 40 km / h) sin km / h js ind the time using the definition for the aerage elocity. (b) It will take him 8.7 min to reach the island. d ae t d t ae 50. km[w0.0 S] t km / h[w0.0 S] t h min t h 1 h t 874. min t 87. min 51

11 Once the direction of the heading (i.e. the elocity of the jet-ski relatie to the water) is found, ector addition can be used to determine the elocity of the jet-ski relatie to the shore (or the resultant ector). The solution can be found from components: x: jsx jrx + rsx or, js cos0.0 jr cosθ + rsx (1) y: jsy jry or, js sin0.0 jr sinθ () Either equation, (1) or () can be soled: rom equation (1): js cos 0. 0 jr cosθ + rsx jr cosθ + rsx ( 40 km / h)cos ( 6.0 km / h) js cos 0. 0 cos km / h or, from equation (): js js js js sin 0. 0 jr sinθ jr sinθ ( km / h)sin17.06 sin 0. 0 sin km / h The magnitude of the elocity calculated here agrees with that calculated from the sine law aboe. 6. rame the Problem - Make a sketch of the motion in the problem. - Vector addition will be used. - The ector sum of the elocity of the cable relatie to the shuttle and the elocity of the shuttle relatie to the space station must be equal to the elocity of the cable relatie to the space station. The elocity of the cable relatie to the space station. Variables and Constants Known Implied Unknown si 1 θ cs 3.0 m s [5 to shuttle] cs x cs y si x si y ci 5

12 Strategy and Draw the ectors on a x-y-coordinate system ( +y coincides with north). y[n] Scale 1.0 cm : 3 m/s ci m s [4.9 to the shuttle] x[e] cs 3.0 m s [E5 S] si 1 m s [E] ind the x-and y-components of the ectors cs and si. si 1 m s [E] cs 3.0 m s [E5 S] ci cs + si cs 3.0 m s [E5 S] θ cs 5 csx cs cos θ cs csx 3.0 m cos 5 s csx 3.0 m s (0.9063) csx.7 m s csy cs sin θ cs csy 3.0 m sin 5 s csy 3.0 m sin(0.46) s cs y 1.7 m csy s Add the components of cs and si to obtain the components of ci. Vector x-component (m/s) y-component (m/s) si cs ci Use the Pythagorean theorem to find the magnitude of ci, and find the angle the resultant makes with the x-axis. ci ( cix ) + ( ciy ) ( ) ci 14.7 m ( ) s m s ci 16.7 m s ci 18.3 m s ci m s ci 15 m s tan θ 1.7 m s 14.7 m s tan θ θ tan θ m s 53

13 Since the x-component is positie and the y-component is negatie the ector and the angle lie in the fourth quadrant. The elocity of the cable relatie to the space station is 15 m/s in a direction 4.9 to the shuttle. Validate The elocity of the cable relatie to the space station is correct since it should be greater than the elocity of the shuttle relatie to the space station alone. The cable possesses the shuttle s elocity as well as its own. Practice Problem Solutions Student Textbook page (a) rame the Problem -Make a diagram of the problem; use a coordinate system with East to the right and north at the top. -Because the ector sum of the three forces is zero, the forces in each dimension must add to zero. -ind the components of the third force that will make the sum of the x and y-components of all the forces equal to zero. The force,, which will make the net force equal to zero. 3 Variables and Constants Identify the Variables Known N[E S] 03 N[W N] Unknown 3 Deelop a Strategy Draw each of the known force ectors with its tail on the origin of the coordinate system and determine the angle each ector makes with the nearest x axis. ind the x-component of each of the force ectors. cos. 0 ( 154 N)(0.9718) 1x 1 1x 1x N The angle is in the 4th quadrant so the x-component is positie. x x cos ( 03 N)(0.7564) x N The angle is in the nd quadrant so the x-component is negatie. 54

14 ind the y-component of each of the force ectors. Sum the x and y-components indiiduallyto find the components of the unknown ector. Use the Pythagorean Theorem to find the magnitude of. 3 Use the tangent function and the magnitudes of the x- and y-components to find the angle θ 3. The angle is in the 4th quadrant so the y-component is negatie. 1x + x + 3x 00. 3x 00. 1x x 3x 00. ( N) ( N) N 3x 1y + y + 3y 00. 3y 00. 1y y 3y 00. ( N) ( N) N 3y 3 ( 3x) + ( 3y) 3 ( N) + ( N) N N 163 N 3 sin. 0 ( 154 N)( ) N 1y 1 1y 1y 3y tanθ 3x. tanθ N N tan θ θ tan ( ) θ θ 58 y y y The angle is in the nd quadrant so the y-component is positie. sin ( 03 N)(0.9616) N The third ector is 163 N[W58 S]. rom a diagram, it is eident that the third ector will be in either the first or third quadrant, so the answer is reasonable. 7. (b) rame the Problem -Same as aboe. Identify the Goal The force,, which will make the net force equal to zero. 3 55

15 Variables and Constants Identify the Variables Known 1 78 N[E1 N] 69 N[W4 S] Deelop a Strategy Draw each of the known force ectors with its tail on the origin of the coordinate system and determine the angle each ector makes with the nearest x axis. ind the x-component of each of the force ectors. ind the y-component of each of the force ectors. Unknown 3 1x N The angle is in the 1st quadrant so the x-component is positie. sin 1. 0 ( 78 N)(0.0791) N 1y 1 1y 1y cos 1. 0 ( 78 N)( ) 1x 1 1x The angle is in the 1st quadrant so the y-component is positie. x N The angle is in the 3rd quadrant so the x-component is negatie. x x y y y cos 4. 0 ( 69 N)( ) sin 4. 0 ( 69 N)( ) N The angle is in the 3rd quadrant so the y-component is positie. Sum the x and y-components indiidually to find the components of the unknown ector. Use the Pythagorean Theorem to find the magnitude of x x 3x 3x 1x x 3x 3x 1y y 3y 3y 1y y 3y 3y 3 ( 3x) + ( 3y) 3 ( N) + ( N) N N 1 N ( N) ( N) N 00. ( N) ( N) N 56

16 Use the tangent function and the magnitudes of the x- and y-components to find the angle θ 3. The third ector is 1 N[W6 N]. rom a diagram, it is eident that the third ector will be in either the second or fourth quadrant, so the answer is reasonable. 7. (c) rame the Problem -Same as aboe. Identify the Goal The force,, which will make the net force equal to zero. 3 3y tanθ 3x. tanθ N N tan θ θ tan ( ) θ θ 58 Variables and Constants Identify the Variables Known 1 48 N[W81 N] 61 N[E63 N] 78 N[E15 S] 3 Unknown 4 Deelop a Strategy Draw each of the known force ectors with its tail on the origin of the coordinate system and determine the angle each ector makes with the nearest x axis. ind the x-component of each of the force ectors. 1x 1 cos x cos x ( 48 N)( ) x ( 61 N)(0.4540) N N 1x The angle is in the nd quadrant so the x-component is negatie. x The angle is in the 1st quadrant so the x-component is positie. 3x 3x 3 cos ( 78 N)(0.9659) 3x N The angle is in the 4th quadrant so the x-component is positie. 57

17 ind the y-component of each of the force ectors. 1y 1 sin y ( 48 N)(0.9877) N 1y The angle is in the nd quadrant so the y-component is positie. y y y sin ( 61 N)(0.8910) N The angle is in the 1st quadrant so the y-component is positie. 3y 3y 3y 3 sin ( 78 N)(0.588) N The angle is in the 4th quadrant so the y-component is negatie. Sum the x and y-components indiidually to find the components of the unknown ector. 1y + y + 3y + 4y 00. 4y 00. 1y y 3y 4y 00. ( N) ( N) ( N) 4y N Use the Pythagorean Theorem to find the magnitude of 4. 4 ( 4x ) + ( 4y ) 4 ( N) + ( N) N N 16 N Use the tangent function and the magnitudes of the x- and y-components to find the angle θ 3. The fourth ector is 16 N[W40 S]. 1x + x + 3x + 4x 00. 4x 00. 1x x 3x 4x 00. ( N) ( N) ( N) N 4x 4 4y tanθ 4x. tanθ N N tan θ θ tan ( ) θ θ 40 rom a diagram, it is eident that the fourth ector will be in the third quadrant to balance the other force ectors, so the answer is reasonable. 58

18 8. rame the Problem - Caitlin acts as the equilibrating force, so the question is similar to the aboe. Identify the Goal The force,, which will make the net force equal to zero. C Variables and Constants Identify the Variables Known A 15 N[N58 E] 18 N[S3 E] B Unknown C Deelop a Strategy Draw each of the known force ectors with its tail on the origin of the coordinate system and determine the angle each ector makes with the nearest x axis. ind the x-component of each of the force ectors. ind the y-component of each of the force ectors. rom the components, the equilibrating ector is in the second quadrant. Sum the x and y-components indiidually to find the components of the unknown ector. Amy s force is 3 to the positie x-axis. Buffy s force is 67 to the positie x-axis. Ax N The angle is in the 1st quadrant so the x-component is positie. Ax Ax 1y N The angle is in the 1st quadrant so the y-component is positie. Ax + Bx + Cx 00. Cx 00. Bx Bx Cx 00. ( N) ( N) N Cx Ay + By + Cy 00. Cy 00. Ay By Cy 00. ( N) ( N) N Cy sin 3 ( 15 N)(0.599) 1y 1 1y A cos 3 ( 15 N)(0.8480) Bx N The angle is in the 4th quadrant so the x-component is positie. Bx Bx y y B cos 67 ( 18 N)(0.3907) sin67 ( 18 N)(0.905) y N The angle is in the 4th quadrant so the y-component is negatie. 59

19 Use the Pythagorean Theorem to find the magnitude of. Use the tangent function and the magnitudes of the x- and y-components to find the angle θ 3. Caitlin exerts a force of N[W4 N]. rom a diagram, it is eident that the third ector will be in either the second or third quadrant, so the answer is reasonable. 9. rame the Problem - The force of graity on the traffic light is balanced by the tension force in the cables. -Draw a free body diagram representing the forces on the traffic light. - The traffic light is in equilibrium and has no net acceleration in either the ertical or horizontal directions. The force that the cables, C Cable C ( Cx) + ( Cy) C ( N) + ( N) C N C N N Cy tanθ Cx N tanθ N tan θ θ tan ( ) θ θ 4, exert on the traffic light to preent if from falling. Variables and Constants Identify the Variables Known Implied Unknown m 65 kg g 981. m/s Cable θ 1 C Deelop a Strategy Apply Newton s second law in the y-direction. Note from the hint that the tension in the two cables is equal. Substitute numerical alues and sole. + mg 0 sin1 + sin1 mg sin1 mg Cable1y Cable1 Cable Cable Cable Cable Cable Cabley Cable mg sin1 ( 65 kg)(9.81 m / s ) (0.079) N 60

20 The tension in the cables is N. (Note that applying Newton s second law in the horizontal direction was unnecessary. The tension in the cables is much greater than the weight of the light (mg 65 kg 9.81 m/s 638 N), as expected, to keep the light steady and resist additional forces due to wind, etc. Therefore, the answer is reasonable. Practice Problem Solutions Student Textbook pages rame the Problem -Sketch a free body diagram of the box on the ramp. -If the component of the force of graity that is parallel to the ramp is greater than the maximum possible friction force, the box will slide down the ramp. Otherwise it will remain motionless. (a) Whether the box will slide down the ramp or remain motionless. (b) The applied force, a, needed to start the box moing, if it was motionless. Variables and Constants Identify the Variables Known Implied Unknown m 45 kg g 981. m/s a θ 1 f µ s 0.4 N f y θ x θ 1 mg Deelop a Strategy ind the x component of the force of graity. gx sinθ g sinθ gx gx gx gx g mg sinθ ( 45 kg)(9.81 m / s )sin N 61

21 ind the maximum possible force of static friction. (a) Since the force of friction is greater than the x-component of the force of graity, the box will remain motionless. Take the difference of the forces to determine the required force down the ramp to get the box to start to slide. An applied force of 15 N is required to get the box to start to slide. µ f(max) s N f(max) f(max) f(max) µ mgcosθ f(max) f(max) s ( 04. )( 45kg)(9.81 m / s )cos N N N gx N gx The ramp has only a shallow tilt, so it is reasonable that the box does not slide. It is also reasonable that a small nudge of 15 N will get it started. 11. rame the Problem -Sketch a free body diagram of the container on the ramp. (It s the same as in the preious question.) - When the component of the force of graity that is parallel to the ramp becomes greater than the maximum possible friction force, the container will begin to slide down the ramp. - When the container begins sliding, Newton s second law can be applied to determine the acceleration down the ramp. (a) The angle, θ, of the ramp at which the container just starts to slide. (b) The acceleration, a x, of the container just after it started to slide. Variables and Constants Identify the Variables Known Implied Unknown m 61 kg g 981. m/s θ µ s 0.37 a x µ k 0.18 Deelop a Strategy ind the x component of the force of graity. ind the maximum possible force of static friction. gx sinθ g sinθ gx gx mg sinθ µ f(max) s N f(max) g µ mgcosθ s 6

22 The container will begin to slide when gx > f(max). So, to find the angle, set the ratio of these to 1 and sole for the angle. (a) The container will begin to slide when the ramp has an angle of 0.3. As soon as the container begins to slide, the coefficient of kinetic friction replaces the coefficient of static friction. ind the acceleration in the x direction (down the ramp) by applying Newton s second law to the components of force in the x direction. The component of the graitational force and the frictional force act in opposite directions so they hae opposite signs. mg sinθ µ ma mg sinθ µ mg cosθ ma a a a a x x x x gx f(max) tanθ 1 µ s tanθ µ tan θ 037. mg sinθ 1 µ mg cosθ θ tan ( 037. ) g(sinθ µ cos θ) 981. m/s (sin cos 03. ) m/s s 1 θ m/s k N x k s k x (b) After the container begins to slide, its acceleration would be 1.7 m/s down the ramp. The angle for the ramp is reasonable (and similar to those gien in preious questions). The alue obtained for the acceleration is much less than the acceleration due to graity (or in free fall), so it is reasonable. 1. rame the Problem -Sketch a free body diagram of the container on the ramp. -By applying Newton s second law to the components of the forces in the x and y-directions, the unknown forces can be determined from the known forces. -Pushing the container at constant elocity implies that the acceleration is zero. The applied force, a, required the push the container up the ramp at a constant elocity. Variables and Constants Identify the Variables Known Implied Unknown m 55 kg g 981. m/s a µ k 0.18 a x 0 θ 33 63

23 N y x a f θ 33 mg Deelop a Strategy Apply Newton s second law in the y-direction. Apply Newton s second law in the x-direction, noting that the acceleration in this direction is zero. Substitute the normal force and expand the terms. Substitute numerical alues and sole. mg cosθ + ma 0 + µ + mg sinθ µ mgcosθ + mg sinθ N gx f a x a f gx a k N a a a a a k mg( µ cosθ + sin θ) k ( 55 kg)(9.81 m / s )( 0. 3cos33 + sin 33 ) N N An applied force of N is required. The force required is slightly less than the weight of the container (mg 55 kg 9.81 m/s 540 N). Because the ramp is relatiely steep, a significant force is required to moe the container along at a constant elocity, so the answer is reasonable. or comparison, moing the container at constant elocity along the ground would require a force, a µmg kg 9.81 m/s 14 N, so, clearly, the steepness of the ramp makes it necessary to apply a much larger force. 13. rame the Problem -Sketch a free body diagram of the crate on the ramp. -By applying Newton s second law to the components of the forces in the x and y-directions, the unknown forces can be determined from the known forces. -In order to start the crate moing, the coefficient of static friction is required. (a) The applied force, a, required the start the crate moing up the ramp, when pushing parallel to the ground. (b) The applied force, a, required the start the crate moing up the ramp, when pushing parallel to the ramp. 64

24 Variables and Constants Identify the Variables Known Implied Unknown m 85 kg g 981. m/s a µ s 0.46 θ 8 y x N θ a f θ mg θ 8 Deelop a Strategy Apply Newton s second law in the y-direction. Note the hint, which says that a component of the applied force is perpendicular to the ramp and thus contributes to the normal force. Apply Newton s second law in the x-direction, noting that the acceleration in this direction is zero. Substitute the normal force and expand the terms. Substitute numerical alues and sole. N gy ay 0 N gy + ay mgcosθ + sinθ (a) An applied force of N is required. When the worker pushes parallel to the ramp, the applied force is parallel to the ramp. So, the terms in a a a a N ax f gx max 0 acosθ µ sn mg sinθ 0 acos θ µ s( mgcosθ + asin θ) mg sinθ 0 acosθ µ smgcosθ µ sasinθ mg sinθ 0 cosθ µ sinθ µ mgcosθ + mg sinθ a s a s a µ smg cosθ + mg sinθ cosθ µ s sinθ ( 046. )( 85kg)(9.81m s )cos 8 + ( 85kg)(9.81m s )sin8 cos 8 + ( 0. 46)sin N N 65

25 the aboe equation (deried from the x-components) that contain a cosθ or a sinθ should be replaced by a and 0, respectiely, because the pushing angle is now zero. Effectiely, this sets the denominator in the aboe a µ smgcosθ + mg sinθ equation to 1. a ( 046. )( 85kg)(9.81m s )cos 8 + ( 85kg)(9.81m s )sin8 Substitute numerical alues a N and sole N (b) To start the crate moing up the ramp by pushing parallel to the ramp requires a force of N. a The units work out to be newtons in each case, as expected. By pushing at a more appropriate angle in the second case, less force is required, so the answer is reasonable. Practice Problem Solutions Student Textbook page Conceptualize the Problem - Begin framing the problem by drawing a free-body diagram of the forces acting on the girl. - The tension of the rope pulls her up, while the force of graity pulls her down. - Let up be positie and down be negatie. - Because the girl accelerates downward, it is expected that the sign of the acceleration will be negatie. - The acceleration can be found by applying Newton s second law. The acceleration, a, of the child as she is being lowered Identify the Variables Known Implied Unknown m 3 kg g 9.81 m/s [up] a t 53 N[up] 66

26 Deelop a Strategy Apply Newton s second law to the free-body diagram. Substitute and sole. The child has an acceleration of about 1.9 m/s [up]. t + g ma t + (m g ) m a t + g m 53 N[up] + (3 kg)( 9.81 m/s [up]) a 3 kg a m/s [up] a 1.9 m/s [up] The child is being lowered from a rope held by her parent, so it is expected that her acceleration will be downward, and much less than the acceleration due to graity, which it is. 15. Solution is similar to Practice Problem 14. The climber has an acceleration of.5 m/s [down]. 16. Conceptualize the Problem -Begin framing the problem by drawing a free-body diagram of the forces acting on the mass on the scale. - The forces acting on the mass are graity ( g ), which is downwards, and the normal force of the scale, which is upwards. -According to Newton s third law, when the mass exerts a force ( MS ) on the scale, it exerts an equal and opposite force ( SM ) on the mass. Therefore, the reading on the scale is the same as the force that the mass exerts on the scale. In the same way, the tension in the hoist rope will also be the same. - The motion is in one direction so let up be positie and down be negatie. -Because the motion is in one dimension, perform calculations with magnitudes only. - Also, because the mass is just starting to moe, the sign of the acceleration will indicate the direction of the motion (the mass is accelerating, not decelerating). -Apply Newton s second law to find the acceleration of the mass. s (a) the direction of motion. (b) the acceleration, a, of the mass. (c) the tension, T, of the hoist rope. Identify the Variables Known Implied Unknown m 10.0 kg g 9.81 m/s MS SM 87 N g T a 67

27 Deelop a Strategy Apply Newton's second law to the free-body diagram. Substitute and sole. (a) The acceleration is negatie, so the motion is downwards. (b) The acceleration of the mass is 1.1 m/s [up]. (c) rom Newton s third law, the tension in the hoist rope will be the same as the force the mass exerts on the scale, or 87 N. + ma SM + ( mg) ma SM SM mg a m 87 N ( kg)(9.81 m / s ) a 10.0 kg a 111. m/s a 11. m/s g Because of the negatie sign of the acceleration (i.e. directed downwards), the reading on the scale is slightly less than it would be if the hoist had zero acceleration. In that case, the scale would read, (10.0 kg)(9.81 m/s ) 98.1 N. This agrees with the experience of riding in an eleator: as it starts traelling downwards, you feel lighter. 17. Solution is similar to Practice Problem 16. The tension of the backpack strap is about N[up]. 18. Solution is similar to Practice Problem 16. The eleator has a maximum acceleration of about 1.8 m/s [up]. Practice Problem Solutions Student Textbook page Conceptualize the Problem - Begin framing the problem by drawing free-body diagrams. Draw one diagram of the system moing as a unit and diagrams of each of the two indiidual objects. - Let the negatie direction point from the centre to the 3.8 kg mass and the positie direction point from the centre to the 4. kg mass. (If the rope was stretched out horizontally, the 3.8 kg mass would be to the left and the 4. kg mass would be to the right.) That is, down is positie for the 4. kg mass. See igure 10.1 in the Student Text, and the Sample Problem, Motion of Connected Objects. - Both objects moe with the same acceleration. The heaier mass will moe downwards and the lighter mass will moe upwards. - The force of graity acts on both objects. - The tension is constant throughout the rope. - The rope exerts a force of equal magnitude and opposite direction on each object. - When you isolate the indiidual objects, the tension in the rope is one of the forces acting on the object. - Newton s second law applies to the combination of the two objects and to each indiidual object. 68

28 Problem Tip Vectors should always be written with both a magnitude and a direction. In problems such as the Atwood machine or letcher s trolley, the directions are usually defined from the centre of the rope and by considering the rope as stretched out. Thus, students may hae two pictures in their minds: the masses hanging on either side of the pulley, and the rope stretched horizontally. Depending on which picture they hae, they may use terms like right and left, or up and down to define the directions of the ectors, which can be confusing for example, talking about graity as acting to the right! Therefore, it is perhaps simplest to define the coordinate system initially, and then to work with the magnitudes of the ectors in the calculations. The acceleration, a, of the two objects The tension, T, in the rope Identify the Variables Known Implied Unknown m kg g 9.81 m/s T m 4. kg a g1 g Deelop a Strategy Apply Newton s second law to the combination of masses to find the acceleration. As the motion can be considered to be in one dimension (if the rope is stretched out), the calculations can be done with magnitudes. Substitute and sole. The acceleration of the masses is about 0.49 m/s. The positie sign indicates the system accelerates in a clockwise direction. Apply Newton s second law to m 1 and sole for tension. Substitute and sole. The tension in the rope is about 39 N. The positie sign indicates the tension on m 1 acts to the right. ma g1 + g (m 1 + m )a m 1 g + m g (m 1 + m )a a (m m 1 )g m 1 + m a (4. kg 3.8 kg)(9.81 m/s ) 4. kg kg a m/s a 0.49 m/s ma g1 + T m 1 a m 1 g + T m 1 a T m 1 g + m 1 a m 1 (g + a) T (3.8 kg)(9.81 m/s m/s ) T N T 39 N You can test your solution by applying Newton s second law to the second mass. 69

29 ma g + T m a T m a m g T m (a g) T (4. kg)( m/s 9.81 m/s ) T N T 39 N The magnitudes of the tensions calculated independently from each of the masses agree. Also, notice that the application of Newton s second law correctly gae the direction for the force on the second mass (to the left). 0. Solution is similar to Practice Problem 19. The mass of the second object is about 14 kg. The tension in the rope is about 75 N[to the right]. 1. Solution is similar to Practice Problem 19. The acceleration of the system is about 1.6 m/s [clockwise]. The mass of the second object is about 6 kg.. Solution is similar to Practice Problem 19. The applied force is about 17 N. 3. Solution is similar to Practice Problem 19. Both gymnasts accelerate upwards at 1.0 m/s. Practice Problem Solutions Student Textbook pages Conceptualize the Problem - Make a simplified diagram of the connected masses and assign forces. - Visualize the light string in a straight configuration. - Sketch free-body diagrams of the forces acting on each object and of the forces acting on the combined objects. - The trolley experiences a horizontal force due to the tension in the string and ertical forces of graity and the normal force, which act opposite to one another. See igure in the Student Text, and the Sample Problem, Connected Objects. - The suspended mass experiences the force of graity and the normal force, which act opposite to one another. - The force causing the acceleration of both masses is the force of graity acting on the suspended mass. - Newton s second law applies to the combined masses and to each indiidual mass. - Let left be the negatie direction and right be the positie direction. (a) The tension, T, in the string when the suspended mass is released (b) The acceleration, a, of the trolley 70

30 Identify the Variables Known Implied Unknown m kg g 9.81 m/s N T m kg a g1 g Deelop a Strategy Apply Newton s second law to the combined masses to find the acceleration. As the motion can be considered to be in one dimension (if the string is stretched out), the calculations can be done with magnitudes (see also the Problem Tip in Practice Problem 19). Substitute and sole. (a) The tension in the string is 3.88 N. Because the sign is positie, the tension acting on m 1 is to the right. (b) The acceleration is about.04 m/s, to the right. Apply Newton s second law to the trolley (mass 1) to find the tension. Use magnitudes in the calculation. net m a g (m 1 + m )a m g (m 1 + m )a m a g m 1 + m a (0.500 kg)(9.81 m/s ) 1.90 kg kg a.0438 m/s a.04 m/s net m a T m 1 a T (1.90 kg)(.0438 m/s ) T 3.88 N The acceleration of the combined masses is less than 9.81 m/s, which is reasonable since only part of the mass is subject to unbalanced graitational forces. The tension can be erified by applying Newton s second law to the suspended mass. net ma T + g m a T m a m g T m (a g) T (0.500 kg)(.0438 m/s 9.81 m/s ) T 3.88 N The negatie sign indicates that the tension on the suspended mass acts upwards. 5. Solution is similar to Practice Problem 4. The glider traels to the end of the track in 0.67 s. 6. rame the Problem -Sketch the apparatus in its correct configuration with forces added. Add a coordinate system. In this case, let the direction to the right be positie. - Then sketch the apparatus with the string in a straight line. -Make free body diagrams of the bodies indiidually. -Only block is experiencing friction with the table. 71

31 -Due to the different masses on either ends of the system, the tensions in strings are not necessarily the same. - Despite the different tensions, the system will moe with uniform acceleration. - The masses on either ends of the system feel the force of graity and the tensions in the strings. -Apply Newton's second law to each free body diagram indiidually in both the x- and y-directions. Examine the set of equations and determine how best to sole for the unknowns. -Once the acceleration is found, use a kinematical equation to find the time it takes the mass to fall. The time interal, t, it takes for the mass, m 3, to reach the ground once the system starts moing. Variables and Constants Identify the Variables Known Implied Unknown m 1 8 g 0.8 kg g 981. m/s a m 615 g kg t m g kg µ k 0.6 d 1.95 m m 1 : T1 m: N m 3 : T T1 f T m 1 g Deelop a Strategy Apply Newton s second law to each free body diagram, in both the x- and y-directions. or m 1. (Only the x-direction is required.) or m. Note the acceleration in the y-direction is zero. or m 3. (Only the x-direction is required.) m g T1 m1g ma 1 (1) x: m a T T1 f or, µ m a T T1 k N N 0 m 3 g y: m g (b) mg 3 T ma 3 (3) (a) 7

32 By inspection, equations (1), (b) and (3) can be rearranged and substituted into equation (a). Collect like terms on each side of the equation. Sole for the acceleration. Substitute numerical alues and sole. µ m a T T1 k N ( mg ma) µ ma 3 3 T1 k N ( mg ma) ( ma+ mg) µ ma k N ( mg ma) ( ma+ mg) µ ( mg) ma k mg ma ma mg µ mg ma k (a) mg mg µ mg ma+ ma+ ma 3 1 k 3 1 ( m + m + m ) a ( m m µ m ) g k ( m3 m1 µ km) g a ( m1+ m + m3) ( kg 0. 8 kg ( 0. 6)( kg) a 981. m /s 0. 8 kg kg kg a m/s a m/s The system accelerates at m/s. Use a kinematics equation that relates the distance, acceleration and time interal to find the time interal. Once the system starts moing, it takes.77 s for the third mass to hit the floor. d 1 at d t a 195 (. m) t m/s t. 773 s t 77. s The units work out properly in both cases. The time interal of almost 3 s to fall about m is ery reasonable. 7. rame the Problem -Sketch the apparatus in its correct configuration with forces added. Add a coordinate system. In this case, let the direction to the right be positie. - Then sketch the apparatus with the string in a straight line. -Make free body diagrams of the bodies indiidually and as one system. -Block 1 is experiencing friction with the ramp. - Both a component of the graitational force and a frictional force on the block are creating a force in the negatie direction. -or m, the tension operates in the negatie direction and the graitational force operates in the positie direction. -ind the acceleration and use the definition of acceleration to find the elocity (or speed). (a) The speed,, of the masses.5 s after they start moing. (b) The tension in the string, T, while they are moing. 73

33 Variables and Constants Identify the Variables Known Implied Unknown m g kg g 981. m/s a m 85 g kg i 0.0 m/s µ k 0.18 T t.5 s N N m 1 : m 1 + m : m : f T f g T g g1 θ g1 θ Deelop a Strategy To find the frictional force, you need the normal force. Apply Newton s second law in the direction perpendicular to the plane. Since the two masses hae the same acceleration, define the system as the combination of the masses. ind the acceleration from Newton s second law in the direction of the string when it is drawn horizontal. Call this the x direction. Substitute in the normal force from the aboe, and expand terms. Sole for the acceleration. Substitute numerical alues and sole. ma 0 N g1 0 N g1 m gcosθ x max gx g1x f ( m1+ m) ax mg m1g sin θ µ kn ( m1+ m) ax mg m1g sin θ µ k( m1g sin θ) ax ( m1+ m) m m1sinθ µ km1sinθ ax g ( m1+ m) kg ( )sin ( 018. )( kg)cos ax 981. m s ( kg kg) a x N m/s 74

34 Use the definition of acceleration to find the elocity. The initial elocity is zero. (a) The speed of the masses after they start to moe will be 0.69 m/s. By examining the free body diagram for the mass m, the tension can be found. Rearrange and sole for the tension. Substitute numerical alues and sole. (b) The tension in the string while they are moing is 0.81 N. ax t ( ) a t ( m / s )(. 5 s) m/s 069. m/s m a m a m g m a m ( g a ) g T x T g x T T T T T x x ( kg)(9.81 m / s m / s ) N 081. N i x The units work out properly in both cases. The speed is equialent to 69 cm/s which seems reasonable considering that the ramp is not steep so the free hanging mass pulls it along quickly. 8. rame the Problem -Sketch the apparatus in its correct configuration with forces added. Add a coordinate system. In this case, let the direction to the right be positie. - Then sketch the apparatus with the string in a straight line. -Make free body diagrams of the bodies indiidually and as one system. -Block 1 is experiencing friction with the ramp. - Both a component of the graitational force and a frictional force on the block are creating a force in the negatie direction. -or m, the tension operates in the negatie direction and the graitational force operates in the positie direction. (a) The force to apply, a, to make the objects start to moe. (b) The acceleration, a, after the objects start to moe. (c) The tension, T, in the string when the objects are moing. Variables and Constants Identify the Variables Known Implied Unknown m 1 75 g 0.75 kg g 981. m/s a m 595 g kg a µ k 0.1 T µ s

35 N N m 1 : m 1 + m : m : f a T f a g T g g1 θ g1 θ Deelop a Strategy To find the frictional force, you need the normal force. Apply Newton s second law in the direction perpendicular to the plane. Sae this equation until it is needed. Since the two masses hae the same acceleration, define the system as the combination of the masses. Note that until the system begins to moe, the acceleration is zero. ma 0 N g1 0 ( kg +(0.75 kg)sin34 + ( 0. 47)( kg)cos34 ) 9. 81m s ind the applied force from a N Newton s second law in the a 091N. direction of the string when it is drawn horizontal. Call this the x direction. Substitute in the normal force from the aboe, and expand terms. Sole for the applied force. Substitute numerical alues and sole. a (a) The applied force is 0.91 N. To calculate the x max acceleration after a + gx g1x f ( m1+ m) ax the system starts to moe, assume the applied 0+mg m1g sin θ µ kn ( m1+ m) ax force is now zero and mg m1g sin θ µ k( m1gcos θ) ( m1+ m) ax proceed as aboe. mg m1g sinθ µ km1gcosθ ax ( m1+ m) m m1 sinθ µ km1 cosθ ax g ( m1+ m) (0.595 kg) ( 01. )( 075. kg)cos34 ( kg)sin34 ax ( kg kg) ax m/s a 087. m/s x N N g1 m gcosθ 1 x max a + gx g1x f ( m1+ m) ax 0 a +mg m1g sinθ µ sn 0 a mg + m1g sin θ + µ s( mg 1 cos θ) ( m + m sinθ + µ m cos θ) g a 1 s 1 76

36 (b) The acceleration of the masses after they start to moe will be 0.87 m/s. By examining the free body diagram for the mass m, the tension can be found. Rearrange and sole for the tension. Substitute numerical alues and sole. (c) The tension in the string while they are moing is 5.3 N. m a m a m g m a m ( g a ) g T x T g x T T T T T x x ( kg)(9.81 m / s m / s ) N 53. N The units work out properly in each case. The ramp is relatiely steep, so a low alue for the acceleration is expected. The acceleration is less than 1/10 of the acceleration due to graity (free fall), which seems reasonable. Practice Problem Solutions Student Textbook page rame the Problem -Sketch the situation, indicating the direction of the force. -Use the equation for torque to determine the magnitude of the torque. The torque, τ, exerted by your biceps muscle. Variables and Constants Identify the Variables Known r 0.05 m 150 N Unknown τ Deelop a Strategy The leer arm is the distance from the hinge of the elbow to where the biceps attaches. Simply apply the equation for torque. The torque is 65 N m. τ r τ ( m)(150 N) τ 65 N m The units are newton metres, the correct unit for torque. 30. rame the Problem -Sketch the situation, indicating the direction of the force. -Use the equation for torque to determine the magnitude of the torque. The torque, τ, exerted by the painter s weight on the ladder. 77

37 Variables and Constants Identify the Variables Known r 0.05 m 150 N Unknown τ Deelop a Strategy The leer arm is the distance from the hinge of the elbow to where the biceps attaches. Simply apply the equation for torque. The torque is 65 N m. τ r τ ( m)(150 N) τ 65 N m The units are newton metres, the correct unit for torque. Practice Problem Solutions Student Textbook page rame the Problem -Sketch the situation, indicating the directions of the forces. - Let the axis of rotation be the fulcrum. This will simplify the calculations (compared to choosing the bolt as the axis of rotation). -Use the equation for rotational equilibrium to find the force the bolt exerts on the board. The force, bolt, the bolt must exert on the diing board to hold it in place. Variables and Constants Identify the Variables Known Implied Unknown m dier 54 kg g 9.81 m/s bolt m board 5 kg l 3.8 m l fulcrum 1.3 m Dier Midpoint of board fulcrum Bolt 0.6 m m fulcrum 1.3 m bolt Deelop a Strategy Begin by writing the equation for rotational equilibrium, using the fulcrum as the axis. 78