Dragging a Crate

Question:

A 500lb crate is initially at rest on the floor of a warehouse. Using a rope with tensile strength of 200lb, attached to the crate and leading upward at a 30 degree angle, what is the maximum acceleration the crate can have if the coefficient of static friction between the crate and the floor is 0.4 and the coefficient of kinetic friction is 0.3?

Answer:

The acceleration of the crate along the horizontal is given by the net horizontal force divided by the mass of the crate. The mass of the crate is its weight divided by the acceleration of gravity, in English units the mass is then 500lb/32.2f/s/s=15.53 slugs.

It would take a force of 250lb in the upward direction at the point of attachment of the rope to the crate to lift one end off the floor. The total tensile strength of the rope is only 200lb so there will be no tilting to worry about.

The force of static friction is the static friction coefficient times the weight of the crate minus the lifting force from the tow rope, or 0.4*(500-T*sin(30)), where T is the tension in the tow rope. The horizontal force available is the the T*cos(30). Using the max strength of the rope for T we can see if there is enough horizontal force available to break the static friction forces and get the box moving. Force available is 200*0.866=173.2lb. Force required is 0.4*(500-200*0.5)=160lb so we can get the crate moving.

Once moving the force required to keep it moving is 0.3*(500-200*0.5)=120lb. The force available is 173.2lb. the difference, 53.2lb, goes into accelerating the crate. The acceleration is 53.2lb/15.53 slugs=3.43f/s/s

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