Question

A 12-kg crate rests on a horizontal surface and a boy pulls onit with a force that is 30° below the horizontal. If thecoefficient of static friction is 0.40, the minimum magnitude forcehe needs to start the crate moving is:

O 44 N
O 47 N
O 54 N
O 56 N
O 71 N

Answer 1

The minimum magnitude force the boy needs to start the crate moving is 47 N. To find the minimum force to move the crate, calculate the weight of the crate and subtract the vertical component of the applied force to find the normal force. The maximum static friction force can then be calculated using the coefficient of static friction. The required force is the horizontal component of the applied force that overcomes this maximum static friction force.

Step-by-step explanation:

To find the minimum magnitude force needed to start the crate moving, we need to calculate the maximum force of static friction. Since the coefficient of static friction is given as 0.40 and the crate has a mass of 12 kg, the normal force acting on the crate is equal to its weight, which is 12 kg x 9.8 m/s² = 117.6 N.

The maximum force of static friction can be calculated using the formula fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force.

Therefore, fs(max) = 0.40 x 117.6 N

= 47.04 N.

Hence, the minimum magnitude force the boy needs to start the crate moving is 47 N, so the correct answer is 47 N.