A mass of 10 kg is suspended by vertically by a rope from the roof. When a horizontal force is applied on the rope at some point the rope deviated at an angle 45 degree at the roof point . If the suspended mass is at equilibrium the magnitude of the force applied is [g = 10m/s]
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There are two cases:-
In first case body of mass 10 Kg is suspended but in Second case body of mass 10 Kg is deviated at an angle of 45°.
So,
=> Case 1 = Case 2
=> F CosØ = mg CosØ
=> F Cos 45 = mg Cos 45
=> F = (10 x 10) Kg m/s^2 [ Both Cos 45 cancel eachother].
=> F = 100 N
In first case body of mass 10 Kg is suspended but in Second case body of mass 10 Kg is deviated at an angle of 45°.
So,
=> Case 1 = Case 2
=> F CosØ = mg CosØ
=> F Cos 45 = mg Cos 45
=> F = (10 x 10) Kg m/s^2 [ Both Cos 45 cancel eachother].
=> F = 100 N
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Answer: 100N
Explanation:
now if we make an FBD of the deviated system which is at equilibrium
we get,
Tsin45=mg,
so here T=100*2^1/2 (100 root 2)
now in horizontal direction,
F=Tcos45,
F=100N AND THATS THE ANSWER
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