Physics, Friction, and inclined planes?

Question

A block weighing 75.0 N rest on a plane inclined at 25.0 degrees to the horizontal. A force F is applied to the object at 40.0 degrees to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and plane are repectively, 0.363 and 0.156
a) what is the min value of F that will prevent the block from slipping down the plane?
b) What is the min value of F that will start the block moving up the plane
c) What value of F will move the block up the plane with a constant velocity



Any help with this would be great. Don't have to give answer ( would be nice to see how it is worked out). Or just a helpful push in the right direction. Thanks alot

Answer 1

First, let's tackle the free body diagram. It's best to resolve the outside force of F parallel and perpendicular to the plane as opposed to using it's angle measure versus the horizontal.

In the x-direction (parallel with the plane) we have the following forces

(-75sin(25)+Fcos(15)-u(75cos(25)-Fsin(15)) (I)

In the y-direction

75cos(25)-Fsin(15) (II)

If you set (I) equal to 0 and use the static friction coef. that should be the stay still force

If you continue and increase the force slightly, the block will move still using the static friction value.

To get constant velocity, switch to the kinectic friction value.. This assumes F acts through the center of gravity.

Answer 2

Case a million : no friction . Conservation of energy. M g h = (0.5 of) m v^2 ---> v^2 = 2 g h. because of the reality that 20*sin30° = 10 m the gadgets will attain the floor with an identical velocity. Case 2: artwork of frictional tension = - 0.2 m g cos30 *L Then: Wf = (0.5)mV^2 - mgL sin30° V^2 = 2 g L(sin30° - 0.2cos30°) = 128.a million (m/s)^2 V = 11.32 m/s < v = 14 m/s.