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A block weighing 600 N has been subjected to a load system as shown in fig. Work out the frictional force for the block and state whether it is in equilibrium or in motion. What additional force needs to be added to 120 N forces so that the block just moves to the left ?
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Asked by borakaushikkumar | 21 Mar, 2022, 10:15: PM
answered-by-expert Expert Answer
Force 350 N that is pulling the block towards right acts in a direction that makes angle 30o with horizontal.
Let us resolve this 350 N force along horizontal and vertical direction as shown in figure.
 
Hence the reaction force from surface to block , R = ( 600 - 350 sin30 ) N = 425 N
 
Friction force f = μ R = 0.25 × 425  ≈ 106 N
 
Net applied external force acting on block = ( 350 cos30 - 120 ) N  ≈ 183 N
 
Direction of net applied external force acting on block is from left to right
 
Net applied external force is greater than Friction force f . Hence the block is in motion from left to right.
 
------------------------------------
 
The extra force Fex required to move the block towards left should overcome the sum of
net applied external force 183 N and friction force 106 N
 
Fex = ( 183 +106 ) N = 289 N
Answered by Thiyagarajan K | 21 Mar, 2022, 11:23: PM
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