A 61-kg crate is at rest on a ramp inclined at an angle of 15° above the horizon

Question

A 61-kg crate is at rest on a ramp inclined at an angle of 15° above the horizontal, and it's your job to move it up the ramp. Today, you're not feeling strong enough to lift the crate (that would require about 600 N of force), but you could exert up to 425 N of force on it. The ramp is set rather low, so you can't push parallel to the surface of the ramp, and have to settle for a horizontal push. If the coefficient of static friction between the crate and the surface of the ramp is 0.40, will you be able to get the crate moving with a horizontal push, or not?
The setup for your solution should include a complete free-body diagram for the crate, showing all four external forces acting on it. Include coordinate axes, with the x-axis parallel to the ramp's surface and the y-axis perpendicular to it. The two forces not parallel to either coordinate axis should be resolved into their x and y components on the diagram. Examples 11, 12, and 13 in the textbook show what this can look like.

Explanation / Answer

YOU CANNOT, this is why:

F(y) = N - mgcos-Fsin=0 y is the direction perpendicular to the incline

N = mgcos+Fsin = 687.429 N

F(x) = Fcos-mgsin-N = ma(x)

For the crate to move,

Fcos-mgsin > N

Fcos-mgsin = 255.796 N

N = 274.971 N

Therefore, since the friction force is greater, the box WILL NOT MOVE

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