Climbing a Ladder: Calculating Max Height with 800N Person

[Dark Visitor]

Problem: An 8 m, 200 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is .45, and the ladder makes a 60 degree angle with the ground. How far up can an 800 N person climb before the ladder begins to slip?

a. 7.8 m
b. 6.8 m
c. 5.8 m
d. 4.8 mI don't know where to start with this problem. Any help would be appreciated. Please go through each step with detail, because I need to do it myself so I understand what I am doing.

 

[cavalier]

Draw the free body diagram. You should have:
-Normal force from the ground.
-frictional force from the ground
-force of gravity on the ladder
-force of gravity on the person
-normal force from the wall.

Set up your equations so net force is 0. Set up your torque equation using the maximum torque the friction can provide and set it equal to 0. The distance of the person from your point of rotation is your unknown. Solve it and you'll have your answer.

 

[Dark Visitor]

In my free body diagram, I have all of those except gravity on person because I do not know where that goes on the ladder.

I also don't understand how to set up the equations. Am I using the right arm rule, where clockwise is negative, and counter-clockwise is positive? And which one does weight (mg) fall under?

 

[cavalier]

If you set your origin to the right end of the ladder then the weight of the person is counter-clockwise.

 

[Dark Visitor]

Oh, okay. I didn't see that at first.

When I set those equations up, does the force equation break into the total forces acting on the y-axis and the total forces acting on the x-axis?

 

[cavalier]

Yes.

 

[Dark Visitor]

Okay, let me make sure this is right so far:

\sumFx = 0
fs - Nw = 0


\sumFy = 0
Ng - mg = 0

(Nw is the normal force by the wall, Ng is normal force by the ground)

Is this correct so far?

 

[cavalier]

Equation for x direction is correct. Equation for y direction needs the weight of the ladder and the weight of the person, but I only see one weight term.

 

[Dark Visitor]

Well, I get lost at that because how do I know where to place the weight of the person? Or does position not matter for this part?

 

[cavalier]

It doesn't matter for the force equations because linear acceleration does not depend on where the force is located.

Position does matter for your torque equations but in that equation, position is the unknown which you are solving for.

 

[Dark Visitor]

Well I did solve for \sumFy, and the three forces in it. But for \sumFx, there is static friction and normal force of the wall, which I don't know either. All I know is the coefficient of static friction. So how do I find those?

 

[cavalier]

Just before the ladder starts to slip, static friction is maximized. The maximum force of static friction is equal to your y normal force times the coefficient of static friction.

From your x equation you know the frictional force is equal to the x normal force.

 

[Dark Visitor]

I know, but how do I find what the y normal force is?

 

[cavalier]

It's equal to the frictional force and you know the frictional force.

 

[Dark Visitor]

I'm really confused. I don't see how I know the frictional force. To find the frictional force, I use the equation

f_s = \muk n

and I know the coefficient of static friction, but what gets plugged in for n in that equation?

 

[cavalier]

Weight of the person plus weight of the ladder.

 

[Dark Visitor]

so I combine those 2 and plug that number into n?

 

[cavalier]

Yes. It should be clear from your free body diagram that the y normal force minus the combined weight of the ladder and person gives the net force acting in the y-direction. Acceleration is zero so the y-normal force equals the combined weight of ladder and person.

 

[Dark Visitor]

So this is what I got:

\mu_kn - n_w = 0
(.45)(800 N + 200 N) - (450 N) = 0

And then in the y direction, I have:

N_g - m_lg - m_mg = 0
1000 N - 200 N - 800 N = 0

(m_lg is weight of the ladder, m_mg is weight of the man, N_g is normal force by the ground)

 

[cavalier]

Yes. Now get your torque equation set up so you can find the position of the person. It is most convenient to use the right end of the ladder as your origin.

 

[Dark Visitor]

Okay, I think I got it.

N_w + f_s - m_lg - m_mg - N_g = 0

Is that right? And does N_w cancel because that is my point of origin?

 

[cavalier]

Oops. For some reason I thought you did your free body diagram the same way I did. I have my free body diagram set up a bit different so the origin is at the bottom. It doesn't matter in any case. Where you set it won't affect your final answer.

The force of friction and the weight will tend to rotate the ladder in the same direction. Their sign should be the same. Ng will rotate it in the opposite direction. The sign should be the opposite. You also need distances and angles in your equations.

Origin at the bottom.
8N_{w}sin(60)-4m_{l}gsin(30)-xm_{p}gsin(30)=0

Origin at the top.
4m_{l}gsin(30)+xm_{p}gsin(30)+8f_{s}sin(60)-8N_{g}sin(30)=0

If you choose to go with the second equation, your x will be the distance from the top of the ladder. Subtract it from 8 to get the distance from the bottom of the ladder.

 

[Dark Visitor]

I used your top formula, and it worked out to 6.79 m, which 6.8 m is one of the answers. So thank you very much for your help. I understood it afterwards.

 

[matt_crouch]

also if you want to look at it again there are some MIT lectures where the lecturer goes through the equations. very helpful
:)

 

[Dark Visitor]

Thanks. I will keep that in mind. 8)