Question

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s¯². At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Answer 1

Given: The mass of the box is 40 kg placed at a distance of 5 m, the coefficient of static friction between the box and the surface is 0.15 and the constant acceleration of the truck is 2 m/ s 2 .

As the truck is accelerating to the left direction therefore, a pseudo force F acts on the box toward the right direction and the friction force f on the box is towards the left direction.

The free body diagram for the box is shown below.

The pseudo force acting on the box is given as,

F=m a ′

Where, m is the mass of the box and a ′ is the acceleration of the truck.

The friction force acting on the box is given as,

f=μmg

Where, μ is the coefficient of friction and g is the acceleration due to gravity.

Applying the equilibrium condition in the horizontal direction, we get

F−f=ma m a ′ −μmg=ma a= a ′ −μg

By substituting the values in the above equation, we get

a=2−0.15×9.8 =0.53  m/s 2

The time taken by box to fall off the truck is calculated by using second equation of motion.

s=ut+ 1 2 a t 2

Where, u is the initial velocity of the box and t is the time.

By substituting the values in the above equation, we get

5=0×t+ 1 2 ×0.53× t 2 t=4.34 s

Thus, the time taken by the box to fall off the truck is 4.34 s.

The distance covered by the truck in time t can be calculated using second equation of motion.

s ′ = u ′ t+ 1 2 a ′ t 2

Where, u ′ is the initial velocity of the truck and a ′ is the acceleration of the truck.

By substituting the given values in the above equation, we get

s ′ =0×4.34+ 1 2 ×2× ( 4.34 ) 2 ≈19 m

Thus, At a distance of 19 m from the starting point, the box fall off the truck.