Question

If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: Fcosθ−μsN=0 Fsinθ+N−mg=0 In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N).

Answer 1

Answer:

F = (μs mg)/(cosθ + μs sinθ)

Explanation:

Hi!

In order to find F in terms of m, g, θ and μs we need to find N in terms of the same variables.

This can be accomplished by solving the second equation (vertical components) for N:

Fsinθ+N−mg=0

N = mg - Fsinθ

Now we can replace the value of N in the first equation (horizontal components):

Fcosθ − μsN=0

Fcosθ − μs(mg - Fsinθ) = 0

Fcosθ − μs mg + μs Fsinθ = 0

F (cosθ + μs sinθ) =  μs mg

Therefore:

F = (μs mg)/(cosθ + μs sinθ)

Hope this helps!

Answer 2

Answer:

$F= \frac{us*m*g}{cos (\alpha) + us * sen(\alpha ) }$

Explanation:

$F*cos (\alpha )- us*N=0\\F*sen(\alpha )+N-m*g=0\\N=m*g-Fsen(\alpha )\\F*cos (\alpha )- us*(m*g-Fsen(\alpha ))=0\\F*cos (\alpha )- us*m*g+us*Fsen(\alpha )=0\\F*(cos (\alpha )+ us*sen (\alpha ))=us*m*g\\F=\frac{us*m*g}{cos (\alpha )+ us*sen (\alpha )}$