Question

A large 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.

Answer 1

Answer:

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 2

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

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

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.