Question

A 30.7 kg box rests on the floor of an elevator. What is the normal force (in N) on the box if the elevator accelerates downward at 3.34 m/s2?

Answer 1

As stated in problem the lift is acclerating down, the forces acting on the box is , N=normal force in upward direction

Force due to gravity which is mg in downward direction

As we write the expression for the forces acting on the body it is .N- mg (since normal is up so positive and mg is negative because it is down)

The net force Fnet=ma , this net force is equal to the forces acting on body as

Ma= N-mg

Now as we know accleration is in downward therefore the expression can be written as -ma=N-mg

So for normal force it is expressed as N=mg-ma=m(g-a)

Now substituting the value of ,m=30.7kg, g=9.8m/s², a=3.34m/s²

N=m(g-a)

N=30.7(9.8-3.34)

N=198.32N

So this is the required normal force