The total force needed to drag a box at constant speed across a surface with coefficient of kinetic friction $\mu$k is least when the force is applied at an angle$\theta$such that:
A)sin$\theta$= $\mu$k.
B)cos$\theta$ = $\mu$k.
C)tan$\theta$= $\mu$k.
D)cot$\theta$= $\mu$k.
E)sec$\theta$ = $\mu$k.

Question

Quizplus · Accepted Answer

The total force needed to drag the box at constant speed is given by F = mg * μk where μk is the coefficient of kinetic friction and m is the mass of the box. The force is applied at an angle θ with the horizontal. Let the applied force be Fx in the horizontal direction and Fy in the vertical direction. 
Fx = F * cosθ 
Fy = F * sinθ 
Since the box is moving at constant speed, the force in the horizontal direction (Fx) must equal the force of kinetic friction (fk). 
fk = Fx = F * cosθ  
fk = μk * m * g

Solving for F: 
F = fk / cosθ 
= μk * m * g / cosθ

Taking the derivative of F with respect to θ and setting it to zero:

dF/dθ = μk * m * g * sinθ / cos^2θ = 0

sinθ / cos^2θ = 0 
sinθ = 0 
θ = 0 or 180 degrees

Since θ cannot be 0 degrees as the force must act at an angle with the horizontal, the only possible solution is θ = 180 degrees.

Substituting θ = 180 into F = μk * m * g / cosθ 
F = μk * m * g / cos180 
= μk * m * g / (-1) 
= - μk * m * g

The negative sign indicates that the force is acting in the opposite direction of motion, which is expected since the force must overcome the force of friction.

Therefore, the best choice is C) tanθ = μk.

The Total Force Needed to Drag a Box at Constant μ\muμ

The Total Force Needed to Drag a Box at Constant $\mu$