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Grade: 11

                        

A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is 30, the block just begins to slide down the plane, The force of static friction is: a)10 kg.wt , b)98 kg.wt , c)49 kg.wt , d)5 kg.wt (answer), e)15 kg.wt How do you get this answer? Thanks!

4 years ago

Answers : (1)

RANJEET KUMAR
32 Points
							
it would be better if i make a block diagram of the problem but it is not possible for me but still i will try to make you understand. first of all 1 kg.wt means a force that is required to accelerate a mass of 1 kg with an acceleration of approx 9.8 m/s2.
here the block is just going to slide down the wedge..so friction force needed to stop the block is mg*sin30.
since, mass  = 10kg, force required  = 10*g*sin30 newton
                                                      = 5*g  newton
                                                       = 5 kg.wt …............... {since 1 kg.wt = 1*g newton, where g = 9.8 m/s2 }
4 years ago
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