Question

A box of dimensions l and b is kept on a truck moving with an acceleration a. If box does not slide my maximum a acceleration for it to remain in equilibrium w.r.t truck is

Answer 1

• Maximum acceleration for the box does not slide is gb/l.

In this case normal force is start shifting to balance the force (as shown in attached figure)

Suppose normal is shifted x distance from its center.

And from the center pseudo force and weight (mg) is acting.

Remaining forces are normal and frictional force.

Now we consider moving clockwise is positive and moving anticlockwise is negative.

So,

By balancing torque about its center is-

T = R × N

N × x - Fr l/2 = 0, where Fr is frictional force

Fr = ma

N = mg

By putting the values we get

mg x - mal/2 = 0

Maximum value of x cab be = b/2

So for maximum value of x we get maximum acceleration-

mgb/2 - mal/2 = 0

a = gb/l

Hence the maximum acceleration is gb/l

Regards