Question

A 12 kg crate, resting on a horizontal surface, is pulled by a force that is applied at an angle of 30° above the horizontal. Find the minimum force needed to start the crate moving if the coefficient of static friction is 0.40.

Answer 1

Answer:

70.6N

Explanation:

We are given that

Mass of crate=12 kg

$\theta=30^{\circ}$

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

Horizontal component force is equal to friction force

$Fcos\theta=f=\mu N$

Force along vertical direction

$Fsin\theta+mg=N$..(2)

Using equation(2) in equation (1)

$Fcos\theta=\mu(Fsin\theta+mg)$

$Fcos\theta=\mu Fsin\theta+\mu mg$

$\mu mg=Fcos\theta-\mu Fsin\theta=F(cos\theta-\mu sin\theta)$

$F=\frac{\mu mg}{cos\theta-\mu sin\theta}$

Substitute the values then we get

$F=\frac{2\times 0.40\times 9.8}{cos 30-0.40sin 30}$

$F=70.6N$

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