Woman sled question

Question

A woman pulls a sled together with its load of mass m kg. Let g be acceleration due to gravity. If her arm makes an angle theta (vertical) with the ground and the co-efficient of friction (mu) (positive constant), the least Force (F) she must exert to move the force is given by F=mg(mu)/sin (theta) + mucos(theta) where mu= 0.21. Find max and min values for F in Newtons for F 0<theta<pi/2 (greater than or equal)

Answer 1

The statement of the problem contains some ambiguities.

I will restate the problem trying to match your statements as much as I can,

and then solve that problem.

Hope that helps.

A woman pulls a sled together with its load of mass m kg.

Let g be the acceleration due to gravity.

If her arm makes an angle θ with the vertical and

the coefficient of friction is μ=0.21, then

the least Force (F) she must exert to move the sled is given by

F = mgμ/(sin(θ) + μcos(θ)).

Find max and min values in Newtons for F on the interval 0≤θ≤π/2.

The way the question is phrased suggests that F is a function of θ and

all the other quantities are considered constants.

The max and min values are either at a critical point or at domain boundary.

For the critical points we need the derivative.

F(θ) = mgμ/(sin(θ) + μcos(θ))

F'(θ) = mgμ(μsin(θ) - cos(θ))/(sin(θ) + μcos(θ))²

F has critical points where F'(θ) = 0

That is where

μsin(θ) - cos(θ) = 0

cotan(θ) = μ

cotan(θ) = 0.21

θ = 0.434π

Thus in total we have three candidates for max and min points θ = 0, 0.434π, and π/2.

We need to evaluate F(θ) at these three points and compare the results.

F(0) = mgμ/(sin(0) + μcos(0)) = mg

F(0.434π) = mgμ/(sin(0.434π) + μcos(0.434π)) = 0.2mg

F(π/2) = mgμ/(sin(π/2) + μcos(π/2)) = 0.21mg

So the answer is that maximum occurs at θ = 0 and minimum at θ = 0.434π