PHYSICS 121 FALL Homework #3 - Solutions. Problems from Chapter 5: 3E, 7P, 11E, 15E, 34P, 45P

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1 PHYSICS 121 FALL Homework #3 - Solutions Problems from Chapter 5: 3E, 7P, 11E, 15E, 34P, 45P

2 3 We are only concerned with horizontal forces in this problem (gravity plays no direct role) We take East as the +x direction and North as +y This calculation is efficiently implemented on a vector capable calculator, using magnitude-angle notation (with SI units understood) F a = m = (90 0 )+( ) =(29 53 ) 30 Therefore, the acceleration has a magnitude of 29 m/s 2

3 7 We denote the two forces F 1 and F 2 According to Newton s second law, F 1 + F 2 = m a, so F 2 = m a F 1 (a) In unit vector notation F 1 =(200N)îand a = (12 sin 30 m/s 2 )î (12 cos 30 m/s 2 )ĵ= (60m/s 2 )î (104m/s 2 )ĵ Therefore, ( F 2 = (20kg) 60m/s 2) ( î+(20kg) 104m/s 2) ĵ (200N)î (b) The magnitude of F 2 is = ( 32 N)î (21 N)ĵ F 2 = F 2 2x + F 2 2y = ( 32) 2 +( 21) 2 =38N (c) The angle that F 2 makes with the positive x axis is found from tan θ = F 2y /F 2x =21/32 = 0656 Consequently, the angle is either 33 or = 213 Since both the x and y components are negative, the correct result is 213

4 11 We apply Eq 5-12 (a) The mass is m = W/g =(22N)/(98m/s 2 )=22kg At a place where g =49m/s 2, the mass is still 22 kg but the gravitational force is F g = mg =(22kg)(49m/s 2 )=11N (b) As noted, m =22 kg (c) At a place where g = 0 the gravitational force is zero (d) The mass is still 22kg

5 15 We note that the free-body diagram is shown in Fig 5-18 of the text (a) Since the acceleration of the block is zero,the components of the Newton s second law equation yield T mg sin θ =0andN mg cos θ = 0 Solving the first equation for the tension in the string, we find T = mg sin θ =(85 kg)(98m/s 2 ) sin 30 =42N (b) We solve the second equation in part (a) for the normal force N: N = mg cos θ =(85 kg)(98m/s 2 ) cos 30 =72N (c) When the string is cut,it no longer exerts a force on the block and the block accelerates The x component of the second law becomes mg sin θ = ma,so the acceleration becomes a = g sin θ = 98 sin 30 = 49 in SI units The negative sign indicates the acceleration is down the plane The magnitude of the acceleration is 49 m/s 2

6 34 First, we consider all the penguins (1 through 4, counting left to right) as one system, to which we apply Newton s second law: F net = (m 1 + m 2 + m 3 + m 4 )a 222 N = (20 kg +15 kg +m kg)a Second, we consider penguins 3 and 4 as one system, for which we have F net = (m 3 + m 4 )a 111 N = (m kg)a We solve these two equations for m 3 to obtain m 3 = 23 kg The solution step can be made a little easier, though, by noting that the net force on penguins 1 and 2 is also 111 N and applying Newton s law to them as a single system to solve first for a

7 45 The free-body diagram is shown below N is the normal force of the plane on the block and m g is the force of gravity on the block We N take the +x direction to be down the incline, in the direction of the acceleration, and the +y direction to be in the direction of the normal force exerted by the incline on the block The x com- (+x) θ ponent of Newton s second law is then mg sin θ = ma; thus,theacceleration is a = g sin m g θ (a) Placing the origin at the bottom of the plane, the kinematic equations (Table 2-1) for motion along the x axis which we will use are v 2 = v ax and v = v 0 + at The block momentarily stops at its highest point, where v = 0; according to the second equation, this occurs at time t = v 0 /a The position where it stops is v 2 0 x = 1 2 ( a ) = 1 ( 350 m/s) 2 2 (98m/s 2 ) sin 320 = 118 m (b) The time is t = v 0 a = v 0 g sin θ = 350 m/s (98m/s 2 ) sin 320 =0674 s (c) That the return-speed is identical to the initial speed is to be expected since there are no dissipative forces in this problem In order to prove this, one approach is to set x = 0 and solve x = v 0 t at2 for the total time (up and back down) t The result is The velocity when it returns is therefore t = 2v 0 a = 2v 0 g sin θ = 2( 350 m/s) (98m/s 2 ) sin 320 =135 s v = v 0 + at = v 0 + gt sin θ = (98)(135) sin 32 =350 m/s

8 PHYSICS 121 FALL Homework #3 - Solutions Problems from Chapter 6: 1E, 9P, 16P, 22P, 39E, 45P

9 1 We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person s push F in the +x direction) Applying Newton s second law to the x and y axes, we obtain F f s,max = ma N mg = 0 respectively The second equation yields the normal force N = mg, whereupon the maximum static friction is found to be (from Eq 6-1) f s,max = µ s mg Thus, the first equation becomes F µ s mg = ma =0 wherewehaveseta =0 to be consistent with the fact that the static friction is still (just barely) able to prevent the bureau from moving (a) With µ s =045 and m =45 kg, the equation above leads to F =198 N To bring the bureau into a state of motion, the person should push with any force greater than this value Rounding to two significant figures, we can therefore say the minimum required push is F = N (b) Replacing m =45 kg with m =28 kg, the reasoning above leads to roughly F = N

10 9 (a) The free-body diagram for the block is shown below F is the applied force, N is the normal force of the wall on the block, f is the force of friction, and m g is the force of gravity To determine if the block falls, we find the magnitude f of the force of friction required to hold it without accelerating and also find the normal force of the wall on the block Wecomparef and µ s N If f < µ s N, the block does f not slide on the wall but if f>µ s N, the block does slide The horizontal component of N F Newton s second law is F N =0,soN = F =12Nand µ s N =(060)(12 N) = 72N The vertical component is f mg =0,sof = mg =50N Since f<µ s N the block does m g not slide (b) Since the block does not move f =50N and N = 12 N The force of the wall on the block is F w = N î+f ĵ= (12 N) î+(50n)ĵ where the axes are as shown on Fig 6-21 of the text

11 16 We choose +x horizontally rightwards and +y upwards and observe that the 15 N force has components F x = F cos θ and F y = F sin θ (a)we apply Newton s second law to the y axis: N F sin θ mg =0 = N = (15)sin 40 +(35)(98)= 44 in SI units With µ k =025, Eq 6-2 leads to f k =11N (b)we apply Newton s second law to the x axis: F cos θ f k = ma = a = (15)cos =014 in SI units (m/s 2 ) Since the result is positive-valued, then the block is accelerating in the +x (rightward)direction

12 22 The free-body diagrams are shown below T is the magnitude of the tension force of the string, f is the magnitude of the force of friction on block A, N is the magnitude of the normal force T T of the plane on block A, m A g N is the force of gravity on body A (where m A = 10 kg), and A m B g is the force of gravity on f B block B θ = 30 is the angle of incline For A we take θ the +x to be uphill and +y to m be in the direction of the normal force; the positive direction A g m B g is chosen downward for block B Since A is moving down the incline, the force of friction is uphill with magnitude f k = µ k N (where µ k =020) Newton s second law leads to T f k + m A g sin θ = m A a =0 N m A g cos θ = 0 m B g T = m B a =0 for the two bodies (where a = 0 is a consequence of the velocity being constant) We solve these for the mass of block B m B = m A (sin θ µ k cos θ) =33 kg

13 39 The magnitude of the acceleration of the cyclist as it rounds the curve is given by v 2 /R, wherev is the speed of the cyclist and R is the radius of the curve Since the road is horizontal, only the frictional force of the road on the tires makes this acceleration possible The horizontal component of Newton s second law is f = mv 2 /R IfN is the normal force of the road on the bicycle and m is the mass of the bicycle and rider, the vertical component of Newton s second law leads to N = mg Thus, using Eq 6-1, the maximum value of static friction is f s,max = µ s N = µ s mg If the bicycle does not slip, f µ s mg This means v 2 R µ sg = R v2 µ s g Consequently, the minimum radius with which a cyclist moving at 29 km/h =81 m/s can round the curve without slipping is R min = v2 µ s g = 81 2 (032)(98) =21 m

14 45 The free-body diagram (for the airplane of mass m) is shown below We note that F l is the force of aerodynamic lift and a points rightwards in the figure We also note that a = v 2 /R where v = 480km/h = 133 m/s Applying Newton s law to the axes of the problem (+x rightward and +y upward) we obtain F l sin θ = m v2 R F l cos θ = mg where θ = 40 Eliminating mass from these equations leads to θ F l tan θ = v2 gr which yields R = v 2 /g tan θ = m m g

15 PHYSICS 121 FALL Homework #3 - Solutions Additional Problems: Drag =? m = 2000 kg Lift =? Force of Gravity Thrust 5000N I The airplane shown above flies with constant velocity in the horizontal direction There are four forces acting on the airplane The thrust force and drag force act in the horizontal direction The force of gravity and the lift force act in the vertical direction a If the thrust force has a magnitude of 5000N, what is the magnitude of drag force? Since the airplane flies at constant velocity the net force on the airplane must be zero Consequently both the horizontal component and the vertical component of the net force must be zero F netx = 0 = Thrust - Drag Consequently, Drag = Thrust = 5000N b If the mass of the airplane is 2000kg, what is the magnitude of the lift force? F nety = 0 = Lift - mg So, Lift = mg = 2000kg x 98 m/s 2 = 19600N or 20000N

16 50 kg n = 40N II A normal force with a magnitude of 40N pushes on a 50 kg mass vertically upward as shown above The only other force acting on the mass is the force of gravity a What is the magnitude of the acceleration of the 50 kg mass? F nety = n - mg = 40N - 50kg(98m/s2) = -9N F netx = 0 F net = 9N a = F net /m = 9N/50kg = 196 m/s 2 = 2 m/s 2 b What is the direction of the acceleration of the 50 kg mass? The direction of the acceleration is the same as the direction of the net force The net force is directed vertically down or - y direction

Easy. P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., f n F F g. (a) 75.0 N N N N (b) ma y.

Easy. P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., f n F F g. (a) 75.0 N N N N (b) ma y. Chapter 5 Homework Solutions Easy P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., (a) f n F F g s k 75.0 N 25.09.80 N 0.306 60.0 N 25.09.80 N 0.245 ANS. FIG. P5.3 P5.4 F y ma y : n mg 0 f s