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CBSE Class 12-science Answered

A wooden block weighing 50N can be just moved on a horizontal plane by a horizontal force of 20 N. What force inclined at 30 degree with the horizontal will be required just to move the block, if (a) force is pull, (b) force is push?
Asked by donehacking97 | 20 Dec, 2020, 05:32: AM
answered-by-expert Expert Answer
Case of Horizontal force :-
 
From Free body diagram of Horizontal force case , we see that normal reaction force R = mg  = 50 N
 
Friction force = μ R = μ 50 N 
 
If 20 N horizontal force just able to move the block, then friction force μ 50 = 20 N  or  kinetic friction coefficient μ = 0.4 
 
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Case of Pushing force :-
 
Pushing force F is applied at an angle 30o to horizontal as shown in free body diagram.
 
This pushing force F is resolved into two components , (i) horizontal component ( F Cos30 ) that is pushing the block in horizontal direction, 
and (ii) vertical component ( F sin30 ) acting normal to the contact surface in the direction of weight .
 
Hence we get Normal reaction force , R = ( mg + F sin30 )
 
Friction force = μ R = μ ( mg + F sin30 )
 
To move the block , horizontal component of applied force should be greater than or just equals the the friction force
 
F Cos30  = μ ( mg + F sin30 )
 
F ( Cos30 - μ sin30 ) = ( μ m g )
 
F  =  ( μ m g ) / ( Cos30 - μ sin30 )
 
By substituting μ = 0.4 , mg = 50 N in above equation we get , F = 30 N
 
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Case of Pulling force :-
 
Pulling force F is applied at an angle 30o to horizontal as shown in free body diagram.
 
This pulling force F is resolved into two components , (i) horizontal component ( F Cos30 ) that is pulling the block in horizontal direction, 
and (ii) vertical component ( F sin30 ) acting normal to the contact surface in opposite direction of weight .
 
Hence we get Normal reaction force , R = ( mg - F sin30 )
 
Friction force = μ R = μ ( mg - F sin30 )
 
To move the block , horizontal component of applied force should be greater than or just equals the the friction force
 
F Cos30  = μ ( mg - F sin30 )
 
F ( Cos30 + μ sin30 ) = ( μ m g )
 
F  =  ( μ m g ) / ( Cos30 + μ sin30 )
 
By substituting μ = 0.4 , mg = 50 N in above equation we get , F = 18.76 N
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