Consider a crate of mass 5.0 kgresting on the floor. A man pushing along a line making an angle of 15° with the plane of the floor is barely able to overcome the friction (coefficient µ_k=0.45) and slide the crate with constant speed. The minimum force necessary to produce this motion is A) 0 N. B) 23 N. C) 46 N. D) infinite.

Question

Quizplus · Accepted Answer

The minimum force necessary to overcome friction and move the crate at a constant speed can be found using the formula $F = \mu_k \cdot N$, where $F$ is the force of friction, $\mu_k$ is the coefficient of kinetic friction, and $N$ is the normal force. The normal force can be calculated as $N = mg \cos(\theta) + F \sin(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $\theta$ is the angle of the applied force with the horizontal. However, since the question asks for the minimum force to overcome static friction and move the crate with constant speed, we use the static friction coefficient $\mu_s = 0.45$, and the force needed is directly related to the normal force and the friction coefficient. The calculation involves balancing the forces to find the minimum force that overcomes the static friction without accelerating the crate, which typically involves solving for the force that equals the maximum static friction force. Given the parameters, the correct calculation leads to a specific value, which in this case is 23 N, indicating the precise force needed to initiate and maintain movement at a constant speed without further acceleration or deceleration.

Consider a Crate of Mass 5