F = mg(μ - sin θ)/(cos θ + μ sin θ)
We want to find the maximum and minimum values of F for 0≤θ≤π/2.
First, let's find the critical points of F by taking the derivative with respect to θ and setting it equal to zero.
dF/dθ = mg[-(1 + μ^2)sin θ cos θ + cos^2 θ]/(cos θ + μsin θ)^2 = 0
To solve for θ, we need to simplify this equation. First, we can notice that if the numerator is zero, the expression will be zero, so we can focus on the numerator:
-(1 + μ^2)sin θ cos θ + cos^2 θ = 0
Rewriting in terms of sin^2 and cos^2 :
cos^2 θ(1 - (1 + μ^2)sin θ) = 0
This gives us two cases. First, consider when cos^2 θ = 0. From this, we can deduce that θ = π/2.
Now, we need to find the values of θ that satisfy:
1 - (1 + μ^2)sin θ = 0
sin θ = 1/(1 + μ^2) = 1/(1 + 0.15^2) = 1/1.0225
θ = arcsin(1/1.0225) ≈ 0.304
Now that we have the critical points, let's find the values of F at these points and the endpoints, θ=0, π/2:
F(0) = mg(0.15 - 0)/(1 + 0) = 0.15mg
F(0.304) = mg(0.15 - sin 0.304)/(cos 0.304 + 0.15 sin 0.304) ≈ mg(0.15 - 0.2997)/(0.9542 + 0.15*0.2997) = 0.150mg
F(π/2) = mg(0.15 - 1)/(0 + 0.15) = -5.6667mg
The minimum force required is at θ=π/2, with F = -5.6667mg.
The maximum force required is at θ=0.304, with F = 0.150mg.
A woman pulls a sled which, together with its load, has a mass of m kg. If her arm makes an angle of θ with her body (assumed vertical) and the coefficient of friction (a positive constant) is μ, the least force, F, she must exert to move the sled is given by
If μ=0.15, find the maximum and minimum values of F for 0≤θ≤π/2. Give answers as multiples of mg.
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