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Pls provide the solution of question attached in image attached
one year ago

Arun
23331 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
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions