Please help: normal force, coefficient of friction

[satishinamdar]

Dear Sir, Kindly help me to solve.

Ladder of length 10 mtr. mass 16 kg rests against a wall at 37 degrees. A man weighing 60 kg is standing 8 meters from the lower end on the ladder. Find the normal force, the force of friction, and minimum coefficient of friction for a man to work safely.

I could get normal force as 745 N, but I could not get correct answer of 412 N for friction force. Please help to answer and explain the steps.

Regards

 

[daon]

Dear Sir, Kindly help me to solve.

Ladder of length 10 mtr. mass 16 kg rests against a wall at 37 degrees. A man weighing 60 kg is standing 8 meters from the lower end on the ladder. Find the normal force, the force of friction, and minimum coefficient of friction for a man to work safely.

I could get normal force as 745 N, but I could not get correct answer of 412 N for friction force. Please help to answer and explain the steps.

Regards

In order for the system not to accelerate we need:

\(\displaystyle \L \sum \tau = 0\)

In our case, we want the force of friction to equal to the force of the ladder on the wall. The force of the ladder will be denoted (normal from the wall) \(\displaystyle N_l\).

Choose the bottom of the ladder where it touches the ground to be the pivot point.

Consider that the mass of the ladder is all at its center of mass \(\displaystyle \frac{10}{2} = 5\)

Also, the man stands at a distance 8 from the pivot point.

Thus our equations are:

\(\displaystyle \L \sum \tau = 0\)
\(\displaystyle \L \Rightarrow (5)(16*9.8)sin(37)+(8)(60*9.8)sin(37) - N_l = 0\)

Therefore
\(\displaystyle \L N_l = 3302N = 10*F*Cos(37) \\ \Rightarrow F = 413N\)

 

[satishinamdar]

Thank you Sir.