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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