Solving Frictional Force on Ladder

[Power of One]

Homework Statement

A ladder 12.0m long weighing 125 N rest against a smooth vertical wall. The bottom of the ladder makes an angel of 67 with the deck. A bucket of paint with a mass of 14kg rests o a rung, 7.00m from the bottom end of the ladder. What is the frictional force exerted on the bottom of the ladder?

Homework Equations

T= F x R x Cos theta

The Attempt at a Solution

Sum of forces x: Nw- fs=0
Sum of forces y= Ng- Fg=0
Sum of torques: Tnw- Tfg=0

Do I have these equations correct? I don't know which angles and lengths get plugged into which? Can someone please help me?

 

[tiny-tim]

Hi Power of One!

First find the normal force.

Then take moments to find the friction force.

 

[kerimek]

I would approach this problem by first identifying the known values and variables. We know the length and weight of the ladder, the angle it makes with the deck, and the mass of the bucket. We also know that the ladder is resting against a smooth vertical wall, meaning there is no friction between the ladder and the wall.

Next, I would draw a free-body diagram to visualize the forces acting on the ladder. We have the weight of the ladder acting downward, the normal force from the ground pushing upward, and the weight of the bucket acting downward at a distance of 7.00m from the bottom end of the ladder.

To find the frictional force on the bottom of the ladder, we can use the equation Fs = μsN, where μs is the coefficient of static friction and N is the normal force. Since the ladder is not moving, we can assume that the static friction force is equal to the force of gravity pulling the ladder down, which is Nw = (125 N)(9.8 m/s^2) = 1225 N.

To find the normal force, we can use trigonometry to find the vertical component of the weight of the ladder. N = Nw x Cos 67 = (1225 N) x (Cos 67) = 474.32 N.

Therefore, the frictional force on the bottom of the ladder is Fs = μsN = μs(474.32 N).

We can also use the equation T = F x R x Cos theta to find the torque on the ladder. The torque due to the weight of the ladder is Tnw = Nw x R x Cos theta = (1225 N) x (12.0 m) x (Cos 67) = 6026.4 Nm. The torque due to the weight of the bucket is Tfg = Fg x R x Cos theta = (14 kg)(9.8 m/s^2) x (7.00 m) x (Cos 67) = 644.6 Nm.

Since the ladder is in equilibrium, the sum of the torques must be equal to zero. Therefore, Tnw - Tfg = 0, which means that the frictional torque (Tfg) is equal to the normal torque (Tnw).

In conclusion, the frictional force on the bottom of the ladder is equal to the normal