
A block weighing $ W $ is held against a vertical wall by applying a horizontal force $ F $ . The minimum value of $ F $ needed to hold the block is if $ \mu < 1 $
(A) Equal to $ W $
(B) Less than $ W $
(C) Greater than $ W $
(D) Data is insufficient
Answer
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Hint : The reaction force of the wall provides minimum force required to hold the block in contact with the wall. The frictional force is equal to the product of coefficient of friction and normal reaction force.
Complete step by step answer:
It is given that a block of weight $ W $ is held against a vertical wall by applying a horizontal force $ F $ .
This implies that the value of applied force $ F $ should be greater than or equal to the reaction force applied by the wall. Otherwise, the block would have fallen down if the value of applied force was less than the reaction force applied by the wall. Thus, the force which is acting perpendicular to the two surfaces in contact with each other i.e., the force applied by the wall on the block is known as normal reaction force and is represented by $ N $ .
This frictional force provides the normal reaction force between the wall and block.
The required frictional force $ {f_N} = \mu N $ where $ {f_N} $ is the force of friction, $ \mu $ is the coefficient of friction and $ N $ is the normal reaction force.
$ \Rightarrow \mu R = W $ , where $ R $ is the reaction force.
$ \therefore $ The force required to hold the block in place should be greater than or equal to the weight of the block i.e., $ {f_N} \geqslant W $ where $ {f_N} $ is the force of friction and $ W $ is the weight of the block.
Let $ F $ be the minimum required force.
Thus, $ W = \mu F $ where
$ \Rightarrow F = \dfrac{W}{\mu } $
As $ \mu < 1 $ , thus $ F > W $ .
$ \therefore \mu F \geqslant W $ .
So, the correct answer is Option C.
Note:
It is the upward friction which holds the mass from falling downward as we know that the maximum value of friction is $ {f_N} = \mu N $ where $ {f_N} $ is the force of friction, $ \mu $ is the coefficient of friction and $ N $ is the normal reaction force.
Normal reaction force $ N $ is equal to the horizontal force because horizontal force is balanced due to no motion in horizontal direction.
The minimum force required is equal to $ W $ when the value of $ \mu = 1 $ .
Complete step by step answer:
It is given that a block of weight $ W $ is held against a vertical wall by applying a horizontal force $ F $ .
This implies that the value of applied force $ F $ should be greater than or equal to the reaction force applied by the wall. Otherwise, the block would have fallen down if the value of applied force was less than the reaction force applied by the wall. Thus, the force which is acting perpendicular to the two surfaces in contact with each other i.e., the force applied by the wall on the block is known as normal reaction force and is represented by $ N $ .
This frictional force provides the normal reaction force between the wall and block.
The required frictional force $ {f_N} = \mu N $ where $ {f_N} $ is the force of friction, $ \mu $ is the coefficient of friction and $ N $ is the normal reaction force.
$ \Rightarrow \mu R = W $ , where $ R $ is the reaction force.
$ \therefore $ The force required to hold the block in place should be greater than or equal to the weight of the block i.e., $ {f_N} \geqslant W $ where $ {f_N} $ is the force of friction and $ W $ is the weight of the block.
Let $ F $ be the minimum required force.
Thus, $ W = \mu F $ where
$ \Rightarrow F = \dfrac{W}{\mu } $
As $ \mu < 1 $ , thus $ F > W $ .
$ \therefore \mu F \geqslant W $ .
So, the correct answer is Option C.
Note:
It is the upward friction which holds the mass from falling downward as we know that the maximum value of friction is $ {f_N} = \mu N $ where $ {f_N} $ is the force of friction, $ \mu $ is the coefficient of friction and $ N $ is the normal reaction force.
Normal reaction force $ N $ is equal to the horizontal force because horizontal force is balanced due to no motion in horizontal direction.
The minimum force required is equal to $ W $ when the value of $ \mu = 1 $ .
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