- #1

- 10

- 0

**A 300 kg box rests on a platform attached to a forklift. Starting from rest at time t=0, the box is lowered with a downward acceleration of 1.5 m/s^2.**

1)Determine the upward force exerted by the horizontal platform on the box as it is lowered.

At time t=0 the forklift also begins to move forward with an acceleration of 2m/s^2 while lowering the box as described above. The box does not slip or tip over.

2) Determine the frictional force on the box.

3) Given that the box does not slip, determine the minimum possible coefficient of friction between the box and the platform.

4) Detmine an equation for the path of the box that expresses y as a function of x (and not of t) assuming that at time t=0, the box has a horizontal position x=0 and a vertical position of y=2m above the ground with zero velocity.

1)Determine the upward force exerted by the horizontal platform on the box as it is lowered.

At time t=0 the forklift also begins to move forward with an acceleration of 2m/s^2 while lowering the box as described above. The box does not slip or tip over.

2) Determine the frictional force on the box.

3) Given that the box does not slip, determine the minimum possible coefficient of friction between the box and the platform.

4) Detmine an equation for the path of the box that expresses y as a function of x (and not of t) assuming that at time t=0, the box has a horizontal position x=0 and a vertical position of y=2m above the ground with zero velocity.

## The Attempt at a Solution

I think I got #1 easily enough. I set up a Fnet=m*a equation and found the upward force=2490N.

#2 is giving me a lot of problems. I cant seem to draw the FBD for this setup. Is there a force acting on the box from the tractor as well as a force due to friction? How can we solve for fricitional force if we dont know the force from the tractor? HELP!

#3 I'll be able to solve for coefficient of friction once I know the force of friction.

#4 NO IDEA WHERE TO BEGIN.