A12 kg crate, resting on a horizontal surface, is pulled by a force that is applied at an angle of 30â° above the horizontal. find the minimum force needed to start the crate moving if the coefficient of static friction is 0.40.

70.6N

Explanation:

We are given that

Mass of crate=12 kg

Coefficient of static friction=[tex\mu=[/tex]0.40

Horizontal component force is equal to friction force

Force along vertical direction

..(2)

Using equation(2) in equation (1)

Substitute the values then we get

Hence, the minimum force needed to start the crate moving=70.6N

1= 26.17

2= 1.17

3=30.86

4=5.86

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