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Grade: 12
        
Pls provide the solution of question attached in image attached
6 months ago

Answers : (1)

Arun
20881 Points
							
 
figure shows the different forces acting on the block, when it is prevented from falling or pushed up. 
when the applied force Fh is ment for preventing the block from falling, friction force act against movement, 
hence friction force acts upward and it is against the direction of pulling force mg×sin45.
 
F= mg×sin45 - μR = mg×sin45 - μ×mg×cos45 = ( mg/√2 )×(1-μ) .................(1)
 
where R is the normal reaction force.
 
when the applied force Fu is ment for pushing the block up, friction force act against movement, 
hence friction force acts downward and it is in the same direction of pulling force mg×sin45.
 
F= mg×sin45 + μR = mg×sin45 + μ×mg×cos45 = ( mg/√2 )×(1+μ) .................(2)
 
It is given that, Fu = 3×Fh  .............(3)
From (1), (2) and (3) we can write,    ( mg/√2 )×(1+μ) = 3×( mg/√2 )×(1-μ) or μ = 0.5
 
N = 10×μ = 10×0.5 = 5
6 months ago
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