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Forces applied at an Angle & Inclined Planes

Forces applied at an Angle & Inclined Planes. When forces are applied at angles other than 90 o , we need to resolve the force into its component vectors. Then, we find the net force in each direction, write our force equations ( F net = ma), and solve. Forces applied at an angle.

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Forces applied at an Angle & Inclined Planes

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  1. Forces applied at an Angle & Inclined Planes

  2. When forces are applied at angles other than 90o, we need to resolve the force into its component vectors. Then, we find the net force in each direction, write our force equations (Fnet = ma), and solve. Forces applied at an angle

  3. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. What do we do first?

  4. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. What do we do first? Draw a free-body diagram.

  5. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. Now what?

  6. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. Now what? Resolve F into components

  7. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. What next?

  8. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. Add the forces in each direction. What should the forces equal in the horizontal? In the vertical?

  9. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. Add the forces in each direction. What should the forces equal in the horizontal? Unbalanced Fnet = ma In the vertical? Fnet= 0.

  10. Forces at an angle – We Do Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal. Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. mg = 800 N m = 80 kg F = 100 N Ffr = 40 N Horizontal direction: F cos θ – Ffr = ma 86.6– 40 = 80 a vertical direction: F sin θ + Fn- mg = 0 50 + Fn = 800 Fn= 750 N a = 0.58 m/s2

  11. 1. A box of books weighing 325 N moves with a constant velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find the coefficient of friction between the box and the floor. 2. Two forces act on a 4.5-kg block resting on a frictionless surface as shown. What is the magnitude of the horizontal acceleration of the block? Forces at an angle – You Do

  12. 1. A box of books weighing 325 N moves with a constant velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find the coefficient of friction between the box and the floor. Constant velocity means no acceleration, which means that the forces in ALL directions are balanced. Ff = 425 cos (35.2) = 361.7 N Ff = μsFn Fn = mg + 425 sin (35.2) = 325 N + 425 sin(35.2)Fn = 548.2 N μs = 0.66 Forces at an angle – You Do

  13. 2. Two forces act on a 4.5-kg block resting on a frictionless surface as shown. What is the magnitude of the horizontal acceleration of the block? Forces at an angle – You Do Resolve into components, Add the forces, then find Fnet = ma. 3.7 N + 5.9 N cos 43 = Fnet = 8.3 N a = Fnet / m = 8 .3 N / 4.5 kg = 1.8 m/s2

  14. q INCLINE Draw a free-body diagram for a block accelerating down a ramp.

  15. Fn Ffr mg q Forces on an Inclined Plane Oh no! This will be ugly! We will have to resolve Fn, Ffr, Fnet, and a into components!!! Or … will we? a

  16. Fn Ffr mg q Forces on an Inclined Plane y Solution: Choose a more convenient coordinate system! Make x be parallel to incline and y be perpendicular. Now, only mg has to be resolved into components. Motion and all the other forces will be in the x or y direction. x a

  17. mg Fn Ffr q q q q mg cos mg sin Forces on an Inclined Plane The only force that we have to resolve into components is weight a y x

  18. mg Fn Ffr q q q q mg cos mg sin Forces on an Inclined Plane The only force that we have to resolve into components is weight Notice that the angle for resolving mg is the same as the angle of the incline a y x

  19. Fn Ffr q q q mg cos mg sin Resolve vector mg into two components. Now instead of three forces, we have four forces direction perpendicular tothe incline: Fnet = ma = 0 Fn= mg cos θ

  20. Fn Ffr q q q mg cos mg sin Resolve vector mg into two components. Now instead of three forces, we have four forces WritetheFnetequations directionperpendicular to the incline: Fnet = ma = 0 Fn= mg cos θ force pressing the object into the surface is not full weight mg, but only part of it, So the normal force acting on the object is only part of full weight mg: Fn = mg cosθ

  21. Fn Ffr q q q mg cos mg sin Resolve vector mg into two components. Now instead of three forces, we have four forces directionparalleltothe incline: Fnet = ma Fnet= Ff – mg sin θ = ma The force that causes acceleration downward is only part of the full force of gravity. Greater acceleration the steeper the slope. If the incline = 0, then there is no horizontal movement due to gravity.

  22. Inclined Plane – We do A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice can support up to 550 N. Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of the bear? What’s my strategy??

  23. Inclined Plane – We do A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice can support up to 550 N. Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of the bear? What’s my strategy?? Draw the free-body diagram Choose a coordinate system with x parallel to incline Resolve mg into components Add vectors perpendicular to plane and set Fnet = ma = 0. Add vectors parallel to plane and set Fnet= ma.

  24. Inclined Plane – We do A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice can support up to 550 N. Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of the bear? m = 60 kg θ = 300 = 0.115 g = 10 m/s2 Perpendicular direction: Fnet= ma a = 0 Fn- mg cos θ = 0 Fn= 520 N < 550 N ice can support him, but he should not eat too much Parallel direction: Fnet= ma mg sin θ – Ffr = ma Ffr= μ Fn= 60 N 300 – 60 = 60 a a = 4 m/s2 cute bear is speeding up!!!!

  25. 3. A block weighing 15.0 newtons is on a ramp inclined at 40.0° to the horizontal. A 3.0 Newton force of friction, Ff , acts on the block as it is pulled up the ramp at constant velocity with force F, which is parallel to the ramp. Find F. 4. A 75 kg box slides down a ramp inclined at 25Owith an acceleration of 3.60 m/s2. a) Find the coefficient of friction. b) What acceleration would a 175 kg box have on this ramp? Inclined Plane – You Do

  26. 3. A block weighing 15.0 newtons is on a ramp inclined at 40.0° to the horizontal. A 3.0 Newton force of friction, Ff , acts on the block as it is pulled up the ramp at constant velocity with force F, which is parallel to the ramp. Find F. Constant velocity = no acceleration. This means that forces in parallel to the inclined plane are also balanced. Ff = F F = 12.6 N Inclined Plane – You Do

  27. 4. A 75 kg box slides down a ramp inclined at 25O with an acceleration of 3.60 m/s2. a) Find the coefficient of friction. b) What acceleration would a 175 kg box have on this ramp? Fnet= mg sin 25 – Ff= ma Ff = mg sin 25 – ma = 40.6 N Ff = μsFnμs = Ff / Fn = 40.6 / mgcos25 = 0.06 a = (mg sin 25 – Ff ) / m = Inclined Plane – You Do

  28. Using pictures, words, and equations, describe how to solve an inclined plane force problem. Exit Ticket

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