Frictional Force ( ): The force that occurs when two object rub against one another and always OPPOSES motion. It's not dependent on area of contact.

Transcription

1 Force Push or pull Law Scientific theory that has been proven for many years (can be changed) Newton's 1 st Law (Law of Inertia): Object at rest stays at rest while an object in motion continues in motion of a straight line until acted on by another force 2 nd Law: If you place a force on an object, it will accelerate in the same direction as that of the force F = ma Mass Quantity of inertia possessed Weight Force of gravity pulling down w = mg (measured in N) N = 3 rd Law: For every action there is an equal and opposite reaction Free Body Diagram (FBD) A drawing that isolates one object and shows all of the forces (known and unknown) acting on that body only * You can't push on a rope; it can only be in tension * Forces only act on a body if they touch it (except for weight/gravity) * Weight is always directed straight DOWN Normal Force ( ) "Normal" means perpendicular (to the ground in this case) and the normal force is a force that the GROUND exerts on a body that keeps it from sinking downwards. Frictional Force ( ): he force that occurs when two object rub against one another and always OPPOSES motion. It's not dependent on area of contact. Static Friction: Occurs in non-moving objects Kinetic (Dynamic) Friction: Occurs in moving objects * Static friction is always greater than kinetic friction * hese equations give the maximum possible value of friction * When breaking weight into components, change the direction of the axes

2 * In order for acceleration to be zero (constant velocity), there must be a force of friction Boxes on a Slope: 1) Draw weight SRAIGH DOWN and determine the value 2) Draw the normal force perpendicular to the slope 3) Draw in force of friction if applicable 4) Break the weight into components (on a new diagram): Use the angle measurement of the slope to find the X and Y 5) Write in a new axis with X and Y slanted to the direction of the weight components 6) he box is not accelerating upwards, so the weight can be used to determine the normal force 7) he normal force can then be used to determine the force of friction 8) Using the X component of the weight and force of friction, the acceleration of the box can be found 9) Chapter 3 formulas can then be used with the information given Coefficient of friction =.15 w=(30kg)(9.8m/s 2 ) w=294n 294N 294Cos30 = N F N = 254.6N 294Sin30 = 147N F f = (.15)(254.6N) F f = 38.19N 147N 38.19N = (30kg)(a) a = 3.63 m/s 2

3 Find the distance the box travels after 3 seconds: d = ½ at 2 Vit d = ½ (3.63m/s 2 )(3s) 2 (0m/s)(3s) d = m Half-Atwood Machine: 1) he thing moving horizontally has a Y acceleration of zero and the object moving vertically has an X acceleration of zero 2) Draw and find the weights of the two objects 3) Draw in the tension (direction the rope is pulling) 4) Draw in and find the normal force of the object moving horizontally 5) Draw in force of friction on horizontal object if applicable 6) Make the direction the objects are moving in positive 7) Set up the force equations in the X and Y for the two objects 8) Substitute the equation (set to tension) into the other equation for tension (don't forget to distribute negatives) 9) Solve for acceleration (it is equal for both objects) 10) Plug the acceleration into either equation to solve for the tension of the rope 11) Using the acceleration and given another piece of information, chapter 3 formulas can be used (especially since Vi = 0) F N *Forces should really be drawn on FBD's* F f 25kg w=(25kg)(9.8m/s 2 ) w=245n 15kg w=(15kg)(9.8m/s 2 ) w=147n F f = (.2)(245N) F f = 49N = (25kg)(2.45m/s 2 )49N 49N = (25kg)(a) 147N = (15kg)(a) = N = 25a49N 147N-(25a49N) = (15kg)(a) a = 2.45 m/s 2 Find the distance the 15kg box drops after 3 seconds: d = ½ at 2 Vit d = ½ (2.45m/s 2 )(3s) 2 (0m/s)(3s) d = m

4 Full-Atwood Machine: 1) he objects are both moving vertically and have an X acceleration of zero 2) Draw and find the weights of the two objects 3) Draw in the tension (up) 4) Make the direction the objects are moving in positive (object with more weight usually goes downwards and vice versa) 5) Set up the force equations in the Y for both objects 6) Substitute the equation (set to tension) into the other equation for tension (don't forget to distribute negatives) 7) Solve for acceleration (it is equal on both objects) 8) he tension can then be obtained by plugging the acceleration into one of the equations 9) Chapter 3 formulas can then be used when given another piece of information (especially since Vi = 0) *Forces should really be drawn on FBD's* 7kg 9kg w=(7kg)(9.8m/s 2 ) w=68.6n w=(9kg)(9.8m/s 2 ) w=88.2n = (7kg)(1.225m/s 2 )68.6N 68.6N = (7kg)(a) 88.2N = (9kg)(a) = N = 7a68.6N 88.2N (7a68.6N) = 9a a = m/s 2 If the distance from the ground of the 9kg box is initially 3m, find how long it will take to hit the ground: d = ½ at 2 Vit 3m = ½ (1.225m/s 2 )(t) 2 (0m/s)(t) t = 2.213s

5 Hanging Signs: 1) he sign has an X and Y acceleration of zero since it is stationary 2) Draw in and find the weight of the sign 3) Break the two ropes holding the sign into tension components 4) Using the angles given, create values using sine and cosine 5) Find the forces in the X (don't forget the direction of the tension component) and set to zero 6) Divide by the trig function on both sides to set (and isolate) one tension equal to another 7) Find the forces in the Y and set to zero 8) Plug in equation of one of the tensions from step 6 into the equation for the Y to solve for the value of the other tension 9) Now that you know the numerical value of one of the tensions, plug it into the equation in step 6 to find the other * In order for the sign not to accelerate in any direction, the X components of the ropes must be equal while the Y components must add up to the weight of the sign *Forces should really be drawn on FBD's* 2 Sin20 1 Sin Cos4 2 Cos2 1000kg w=(1000kg)(9.8m/s 2 ) w=9800n 2 =.75( N) 2 Cos20 1 Cos45 = 0 1 Sin45 2 Sin N = 0 2 = N 2 = 1 Cos45/Cos20 1 Sin Sin N = 0 2 = = 9800N 1 = N

6 Multiple Objects Attached to Ropes (moving horizontally): 1) Create one large FBD by adding the masses of the boxes and find acceleration in the X 2) Acceleration in the Y is zero 3) Create individual FBD's and using the acceleration found in step 1, find the tension of the ropes 4) For the final box, there is nothing left to solve for, so a check can be made 200N *Need to show work when finding normal force and weight* 10kg 10kg 20kg ABC = 40kg 40kg F N = 392N 200N 200N = (40kg)(a) a = 5 m/s 2 w=392n F N = 98N F N = 98N F N = 196N 10kg kg 2 20kg 200N w=98n w=98n w=196 1 = (10kg)(5m/s 2 ) -50N 2 = (10kg)(5m/s 2 ) -100N200N = (20kg)(5m/s 2 ) 1 = 50N 2 = 100N 100N=100N Lawnmower (or any other question concerning a diagonal force being exerted on an object):

7 1) he object has a Y acceleration of zero 2) Draw in and find the weight of the object 3) Draw in the normal force 4) Draw in the force of friction if applicable 5) Break the diagonal force into components on a separate diagram 6) Using the angle given, find the X and Y components 7) Using the Y component and weight, sum the forces in the Y and set it to zero to find the normal force 8) Use the normal force to find the value for the force of fraction if given the coefficient of friction 9) Use the force equation in the X's to find acceleration 10) Chapter 3 equations can then be used if other information is given (especially since vi = 0) 10kg F N F f Coefficient of Friction =.2 w=(10kg)(9.8m/s 2 ) w=98n 40Sin30 40Cos30 F f = (.2)(118N) 34.64N-23.6N = (10kg)(a) F N 98N 20N = 0 F f = 23.6N a = m/s 2 F N = 118N Find the velocity of the lawnmower after moving for 3 seconds: Vf = Vi at Vf = 0m/s (1.104m/s 2 )(3s) Vf = m/s