Pls give a detailed solution of this question..

Solution:
(a)
 acceleration of the block =Fcosθm So a =10cos375.1 =10×0.765.1 =1.49m/s2

(b)`To lift the block, the vertical component of the applied force should be equal to its weight. Therefore,

P' sin 370=5.1×10P'=53×51=5×17P'=85 N

(c) Now, Just before the block leaves the contact force, then the applied force n horizontal direction is.


ma=P cos θa=85×cos 375.1=17×45.1a=685.1a=13.33 ms

 



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