Laws of MotionMedium
Question
A force F pushes a block weighing 10 kg against a vertical wall as show in the figure. The coefficient of friction between the block and wall is 0.5. The minimum value of F to start the upward motion of block is [g =10 m/s2]
Options
A.100 N
B.500 N
C.
N
ND.can′t be determined
Solution
N = F cos 37o = 
F and to start upward motion F sin 37o = 10g + μN
F = 100 + (0.5) 
⇒ F = 500 N
F and to start upward motion F sin 37o = 10g + μN F = 100 + (0.5) ⇒ F = 500 NCreate a free account to view solution
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