A block of mass 4kg is placed on a rough horizontal surface with coefficient of friction 1/√3.a force F is applied on the block making an angle theta with horizontal.1) for what value of theta force required to move the block is minimum.2) what is the minimum value of force F required to move the block.3) if the value of F is 4N more than its value calculated in previous problem the acceleration of block will be?
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rameshyadavyadav53:
acceleration options are a)2/√3 b)2√3 c)4//√3 d)1. Force Options are a)20/√3 b)20 c)40//√3.d)40.angles are a)0 b)30 c)45 d)80.
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Answer:
Explanation:
theta = tan^- meu
theta = tan^- 1/root 3
theta = 30
minimum force = mgsin(30)
minimum force = 20
force=24
friction force acting on it = meu(n)
n=mg+fsintheta
n=40-12 = 28
friction force =28/root3
acceleration = 8/4 = 2
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