A block of mass 10 kg is held at rest against a rough vertical wall (μ= 0.5 ) under the action a force F . The minimum value of F required for it  is (g=10 ms-2)



162.6 N

89.7 N

42.7 N

95.2N

pls explain properly

Dear Student ,

Vertically upward component of F = F Cos 30 = ( 31/2/2)F = 0.87F
Normal reaction from the wall is balanced by Horizontal component of F so Normal reaction N = F Sin 30 = F/2
For minimum value of F the frication will be limiting f = 0.5N = 0.25F        (limiting friction  = co-efficient of friction (0.5) times normal)
Hence net upward force = 0.25F + 0.87F = 1.12F
Net downward force = mg = 100 N
For equilibrium Net upward force = Net downward force
1.12F = 100
F = 89.7 N
Option d


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