Question

In: Physics

A 1.02 kg block is kept stationary on a 45 degree slope by applying a horizontal...

A 1.02 kg block is kept stationary on a 45 degree slope by applying a horizontal force F onto it. What is the minimum value of F if the coefficient of static friction is 0.5?

a) 3.33 N

b) 2.56 N

c) 4.25 N

d) 6.40 N

Expert Solution

F_n = normal force

from the force diagram, force equation perpendicular to the incline is given as ::

F_n = F_g cos45 + Fsin45

frictional force is given as

f = F_n

f = (F_g cos45 + Fsin45 ) Eq-1

from the force diagram , force equation parallel to the incline is given as ::

FCos45 + f = F_g Sin45

Using equation 1

FCos45 + (F_g cos45 + Fsin45 ) = F_g Sin45

m = 1.02 kg ,

F_g = 1.02 x 9.8 = 9.996 N

F Cos45 + (0.5) (9.996Cos45 + F Sin45) = 9.996 Sin45

F = 3.33 N

genius_generous answered 1 day ago

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