A block weighing 10kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. If a force acts down at 60° fro m the horizontal, how large can it be without causing the block to move? (g=10ms)
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HEY MATE, HERE IS YOUR ANSWER
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Considering horizontal forces being balanced just before the motion of the block,
Fcos 60=ms*N
F/2=0.5*N
N=weight of block+vertical component of force applied
=100+F sin60
=100+√3F/2
therefore,
F/2=(1/2)*(100+√3F/2)
2F=200+√3F
Therefore,
F=200/(2-√3)
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____________________________
Considering horizontal forces being balanced just before the motion of the block,
Fcos 60=ms*N
F/2=0.5*N
N=weight of block+vertical component of force applied
=100+F sin60
=100+√3F/2
therefore,
F/2=(1/2)*(100+√3F/2)
2F=200+√3F
Therefore,
F=200/(2-√3)
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