Question
Question: Two trolleys 1 and 2 are moving with acceleration \({a_1}\) and \({a_2}\) respectively in the same d...
Two trolleys 1 and 2 are moving with acceleration a1 and a2 respectively in the same direction. A block of mass “ m “ on trolley 1 is in equilibrium from the frame of observer stationary w.r.t trolley 2. The magnitude of friction force on the block due to the trolley is (assume that no horizontal force other than frictional force is acting on the block).
Solution
By the given data, compare the acceleration of the block and the observer and the acceleration of the block and the second trolley. Use the formula of the force given below, substitute the known values, to find the value of the frictional force.
Formula used:
By the Newton’s second law of motion,
F=ma
Where F is the force acting on the block, m is the mass of the block and a is the acceleration of the block.
Complete step by step solution:
It is given that the acceleration of the trolley one is a1.The acceleration of the trolley two is a2.The mass of the block on the trolley one is m.The frame of the observer is in equilibrium with the block.Let the acceleration of block of mass with respect to
the observer on trolley 2 be denoted as abo.
This shows that the acceleration of the block of mass and the observer are the same.
abo=0
ab−ao=0
where ab denotes the acceleration of the block with respect to ground and ao denotes the acceleration of the observer with respect to ground.
Since the observer stands on the trolley two, the acceleration of the observer and the trolley two are the same.
⇒ab=a2--(1)
Let us assume that only frictional force acts on the block. Using the formula of motion,
⇒F=ma
⇒F=mab
Substitute (1) in the above step, we get
⇒F=ma2
Hence the frictional force on the block is ma2.
Note: The frictional force is the force that acts between the two surfaces that keeps the object to remain in place without movement. It has many kinds of use in daily life. We are able to stand only due to friction between the ground and toes.