How Do Forces and Torque Balance in a Ladder Problem with a Firefighter?

[vipertongn]

Homework Statement

A uniform ladder of length L and mass m1 rests against a frictionless wall. The ladder makes an angle θ with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when a firefighter of mass m2 is a distance x from the bottom. (Answer using m_1 for m1, m_2 for m2, theta for θ, g for gravity, and L and x as necessary.)
horizontal

(b) If the ladder is just on the verge of slipping when the firefighter is a distance d from the bottom, what is the coefficient of static friction between ladder and ground?

Homework Equations

Net Fy = 0
Net Fx = 0
Net T = 0

The Attempt at a Solution

I tried n_g-m_1*g-m_2*g as the answer for the vertical force because the ladder has a normal force that is exerted in reaction to the two gravitational forces from the ladder itself and the firefighter. I believe that this isn't the correct way to input it.

for the force for horizontal I know its the frictional force minus the normal force from the wall. however I don't know how to input it using hte variables they want.

 

[tiny-tim]

A uniform ladder of length L and mass m1 rests against a frictionless wall. The ladder makes an angle θ with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when a firefighter of mass m2 is a distance x from the bottom. (Answer using m_1 for m1, m_2 for m2, theta for θ, g for gravity, and L and x as necessary.)
horizontal

(b) If the ladder is just on the verge of slipping when the firefighter is a distance d from the bottom, what is the coefficient of static friction between ladder and ground?

Hi vipertongn!

Yes, the normal force will be equal and opposite to the other vertical forces.

To find the horizontal forces, take moments about some convenient point.

 

[nlightner]

I would approach this problem by first drawing a free-body diagram of the ladder to understand the forces acting on it. From the problem statement, we know that there are three forces acting on the ladder: the weight of the ladder (m1g), the weight of the firefighter (m2g), and the normal force from the wall (N). The ladder is in equilibrium, so the sum of all forces in the x direction and the y direction must be equal to zero.

(a) We can use the equations Net Fy = 0 and Net Fx = 0 to solve for the horizontal and vertical forces. In the y direction, we have:

N - m1g - m2g = 0

Solving for N, we get:

N = m1g + m2g

In the x direction, we have:

Fx = 0

Therefore, the horizontal force is equal to zero.

(b) To determine the coefficient of static friction, we can use the torque equilibrium equation, Net T = 0. The torque due to the weight of the ladder and the firefighter will be counteracted by the torque due to the normal force from the wall. This can be expressed as:

m1gLcosθ + m2gxsinθ = Nd

Where d is the distance from the bottom of the ladder to the firefighter. Solving for N, we get:

N = (m1gLcosθ + m2gxsinθ)/d

The coefficient of static friction (μ) is equal to the ratio of the maximum frictional force (μN) to the normal force (N). Therefore, we can express μ as:

μ = (μN)/N = μN/(m1g + m2g)

Substituting for N, we get:

μ = μ(m1gLcosθ + m2gxsinθ)/(m1g + m2g)

This is the coefficient of static friction between the ladder and the ground when the ladder is on the verge of slipping.