A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Question

3 months ago

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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.

Physics

2 answers:

Yuliya22 [3.3K] · Answer 1 · 2026-02-21T23:57:39+03:00

Answer:

a) F = μk*m*g/(cosθ - μk*sinθ)

b) μs = cotθ

Explanation:

a)

Considering that there is no movement along the y-axis, we establish:

- F*sinθ - m*g + N = 0 (1)

where m*g signifies the weight and N stands for the normal force.

The body is in constant motion along the x-axis, leading us to:

F*cosθ - f = 0 (2)

where f represents friction, given by:

f = μk*N (3)

By combining equations 1, 2, and 3, we have:

F*cosθ = μk*(F*sinθ + m*g)

To isolate F, we rewrite it as:

F*cosθ = μk*F*sinθ + μk*m*g

Thus, we can find F as:

F*(cosθ - μk*sinθ) = μk*m*g

Resulting in:

F = μk*m*g/(cosθ - μk*sinθ)

b)

<pSimilarly, for the static friction scenario, we obtain:

F = μs*m*g/(cosθ - μs*sinθ)

Here, F corresponds to the minimum force necessary to set the crate in motion. This implies that F must exceed zero, setting the condition:

cosθ ≥ μs*sinθ

In the boundary case:

cosθ = μs*sinθ

Thus, the static friction coefficient result is μs = cotθ.

ValentinkaMS [3.4K] · Answer 2 · 2026-02-22T00:00:23+03:00

Answer:

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

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

Explanation:

To address this issue, it’s essential to create a visual representation and a free body diagram (see attached image).

Upon reviewing the diagram and the outlined problem, we observe that the crate maintains a steady velocity. This indicates no acceleration is present, meaning the total forces must equate to zero as per Newton's third law. We can evaluate the forces in both x and y directions. Let's focus on the y-direction analysis:

$\Sigma F_{y}=0$

Three forces act in the y-direction: the weight of the crate, the normal reaction force, and the y-direction force, so our force equation is represented as:

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

Solving for the normal force leads us to:

$N=F_{y}+W$

We recognize that

W=mg

and

$F_{y}=Fsin \theta$

By substitution, we find:

N=F sin θ +mg

Furthermore, we understand that kinetic friction is characterized by:

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

Thus, we can determine kinetic friction by substituting N, resulting in:

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

Next, we evaluate the sum of forces in the x-direction:

$\Sigma F_{x}=0$

This allows us to form our x-direction force sum:

$-f+F_{x}=0$

Applying the known relationships:

$F_{x}=Fcos \theta$

We can now implement our earlier equations into the x-force sum as follows:

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

Now, we can solve for the force by distributing $\mu_{k}$ , yielding:

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

By adding $\mu_{k}mg$ to both sides, we attain:

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

Then we factor out F, resulting in:

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

Finally, dividing both sides by $(cos \theta-\mu_{k} sin \theta)$ , we conclude:

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

This provides the solution for part a.

For part b, we will retain the same free body diagram, but this time the friction coefficient used will be for static friction. By applying the same method as in the earlier part, we derive the equations:

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

and

F cos θ = f

Upon substitution, we produce:

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

This can be manipulated to solve for the static friction coefficient, leading to:

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

This reflects the answer for part b.