Alarge crate with mass m rests on a horizontal floor. the static and kinetic coefficients of friction between the crate and the floor are μs and μk, respectively. a woman pushes downward on the crate at an angle θ below the horizontal with a force f⃗.

a) what is the magnitude of the force vector f⃗ required to keep the crate moving at constant velocity?

express your answer in terms of m, g, θ, and μk.

b) if μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this critical value of μs.

express your answer in terms of θ.

a) F = μk*m*g/(cosθ - μk*sinθ)

b) μs = cotθ

Explanation:

a)

Given that the body is not moving in the y-axis direction, then:

- F*sinθ - m*g + N = 0     (1)

where m*g is the weight of the body and N is the normal force.

Given that the body is moving at constant velocity in the x-axis direction, then:

F*cosθ - f = 0      (2)

where f is the friction, computed as:

f =  μk*N      (3)

Combining equations 1, 2 and 3:

F*cosθ = μk*(F*sinθ + m*g)

Isolating F:

F*cosθ = μk*F*sinθ + μk*m*g

F*(cosθ - μk*sinθ) = μk*m*g

F = μk*m*g/(cosθ - μk*sinθ)

b)

Analogously, for the static case we get:

F = μs*m*g/(cosθ - μs*sinθ)

where now F is the minimum force to move the crate. Notice that F must be positive, so:

cosθ ≥ μs*sinθ

In the limit case:

cosθ = μs*sinθ

μs = cotθ

answer: (431,2)

step-by-step explanation:

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