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.
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.
To find the maximum and minimum values of F, we need to analyze the forces acting on the sled:
1. Weight force (mg): This force acts vertically downward and has a magnitude of mg, where m is the mass of the sled and g is the acceleration due to gravity.
2. Normal force (N): Since the sled is on a horizontal surface, there is an equal and opposite normal force acting vertically upward, which cancels out the weight force.
3. Friction force (f): The friction force acts horizontally opposite to the direction of motion. Its magnitude can be given by f = μN, where μ is the coefficient of friction.
4. Force applied by the woman (F): This force is exerted at an angle θ with respect to the vertical. We need to find the least force required to move the sled, which occurs when the friction force is at its maximum.
Using the above information, we can write the following equations:
Vertical component: N - mg = 0
Horizontal component: F - f = 0
Substituting f = μN, we get:
F - μN = 0
From the vertical component equation, we have:
N = mg
Substituting this into the horizontal component equation, we get:
F - μmg = 0
F = μmg
Therefore, the least force F required to move the sled is μmg.
Given that μ = 0.15 and 0 ≤ θ ≤ π/2, we can calculate the maximum and minimum values of F as follows:
Minimum value of F: When θ = 0, F = μmg = 0.15mg
Maximum value of F: When θ = π/2, F = μmg = 0.15mg
So, for 0 ≤ θ ≤ π/2, the maximum and minimum values of F are both equal to 0.15mg.
To find the maximum and minimum values of F for the given condition, we need to analyze the forces acting on the sled and use Newton's second law of motion. Let's break down the forces involved:
1. The force of gravity (mg): This is the force exerted on the sled due to its mass. It acts vertically downwards.
2. The normal force (N): This is the force exerted by the ground on the sled perpendicular to the surface. It counteracts the vertical component of the gravitational force.
3. The force of friction (f): This is the force opposing the motion of the sled. It acts parallel to the surface and depends on the coefficient of friction (μ).
4. The force exerted by the woman (F): This is the force applied by the woman at an angle θ to the vertical.
Now, let's analyze the forces at different angles θ:
1. When θ = 0: In this case, the force applied by the woman is purely vertical. The force of friction is also zero as there is no horizontal force acting on the sled. Therefore, the minimum force required (F_min) would be equal to the weight of the sled, which is F_min = mg.
2. When θ = π/2: In this case, the force applied by the woman is purely horizontal. The vertical component of the force doesn't contribute to overcoming friction. Therefore, the maximum force required (F_max) is equal to the force of friction, which is given by f = μN. To find N, we need to resolve the weight of the sled into vertical and horizontal components. The vertical component is mg * cos(θ), and the horizontal component is mg * sin(θ). Since the sled is on the verge of tipping over, the horizontal component of the weight is equal to the force of friction, making f = mg * sin(θ) = μN. Solving for N, we get N = mg * sin(θ) / μ. Thus, F_max = f = μN = μ * mg * sin(θ) / μ = mg * sin(θ).
In summary:
- When θ = 0, F_min = mg.
- When θ = π/2, F_max = mg * sin(θ).
Therefore, the maximum and minimum values of F for 0 ≤ θ ≤ π/2, expressed as multiples of mg, are:
- F_min = 1mg.
- F_max = sin(θ)mg.