A block of mass 15kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5ms−2 for 20s and then moves with uniform velocity. Discuss the motion of the block as viewed by,
(a) a stationary observer on the ground. (b) an observer moving with the trolley.
A block of mass 15kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5ms−2 for 20s and then moves with uniform velocity. Discuss the motion of the block as viewed by,
Mass of the block, m = 15 kg
Coefficient of static friction, = 0.18
Acceleration of the trolley, a = 0.5 m/s2
As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 x 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f=μmg
= 0.18×15× 10 = 27 N
∵fl≥Fex⟶Aslongasexternalforceislesserthanlimitingfriction,theblockremainsstationary.
In other words, The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at moving with constant acceleration with respect to ground.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. In this case of a non-inertial frame of reference, The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.
Mass of the block, m = 15 kg
Coefficient of static friction, = 0.18
Acceleration of the trolley, a = 0.5 m/s2
As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 x 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f=μmg
= 0.18×15× 10 = 27 N
∵fl≥Fex⟶Aslongasexternalforceislesserthanlimitingfriction,theblockremainsstationary.
In other words, The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at moving with constant acceleration with respect to ground.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. In this case of a non-inertial frame of reference, The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.
