We consider a man of mass m = 84 kg as shown in the figure below using crutches. The crutches each make an angle of θ = 26degrees with the vertical. Half of the person's weight is supported by the crutches, the other half is supported by the normal forces acting on the soles of the feet. Assuming that the person is at rest, find the magnitude of the force supported by each crutch.

Question

We consider a man of mass m = 84 kg as shown in the figure below using crutches. The crutches each make an angle of θ = 26degrees with the vertical. Half of the person's weight is supported by the crutches, the other half is supported by the normal forces acting on the soles of the feet. Assuming that the person is at rest, find the magnitude of the force supported by each crutch.

Thanks!

Answer 1

Let the force on each crutch be C Newtons.

m=mass of person kg
g=acceleration due to gravity m/s/s
Resolve all forces acting on the person in the vertical direction,
Forces from soles + forces from crutches = weight
(1/2)*mg + 2*C*cos(26°) = mg
C=(1/2)mg/(2*cos(26°))

Answer 2

To find the magnitude of the force supported by each crutch, we'll need to analyze the forces acting on the person and use the principles of static equilibrium.

Let's break down the forces acting on the person:

1. Weight (mg): The weight of the person acts vertically downward with a magnitude of m * g, where m is the mass of the person and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Normal force on the feet (N): Since half of the weight is supported by the normal forces acting on the soles of the feet, each foot supports a normal force of N/2.

3. Force supported by each crutch (F): The crutches make an angle of θ with the vertical. The vertical component of the force supported by each crutch will be F * cos(θ) and the horizontal component will be F * sin(θ).

According to the problem statement, the person is at rest, which means the net force acting on the person is zero.

Now, let's set up the equations:

In the vertical direction:
N/2 + F * cos(θ) = mg

In the horizontal direction:
F * sin(θ) = 0 (since there is no horizontal acceleration)

To solve these equations, we need to substitute known values:

m = 84 kg (mass of the person)
θ = 26 degrees (angle of the crutches with the vertical)
g = 9.8 m/s^2 (acceleration due to gravity)

Now, let's solve for N and F:

From the horizontal equation, F * sin(θ) = 0, we can see that F = 0 since sin(θ) ≠ 0.

Substituting this into the vertical equation:
N/2 + 0 = mg
N = 2mg
N = 2 * 84 kg * 9.8 m/s^2
N ≈ 1640.8 N

Since each foot supports half of the normal force, the magnitude of the force supported by each crutch is N/2:
F = N/2 ≈ 1640.8 N / 2
F ≈ 820.4 N

So, the magnitude of the force supported by each crutch is approximately 820.4 Newtons.