Torque Question: How Far Up a Ladder Can 800N Worker Climb?

[oneplusone]

Homework Statement

A ladder 6 m long and weighs 120 N. It leans against a smooth wall of negligible friction making an angle of 50° with the horizontal. The coefficient of friction between the floor is 0.5. How far up the ladder can an 800 N worker climb before it starts to slip?



I ended up with the equation:

$d*800\sin 40 = 0.5*2000*3*sin 50$

 

[462chevelle]

why do you have sin(40) and sin(50). would one of those not be cos?

 

[oneplusone]

But for torque, isn't it:
distance * force * sin x

 

[462chevelle]

well. if you draw the FBD the 40 degree angle is in one direction and the 50 is in the other direction.

 

[darkxponent]

Homework Statement

A ladder 6 m long and weighs 120 N. It leans against a smooth wall of negligible friction making an angle of 50° with the horizontal. The coefficient of friction between the floor is 0.5. How far up the ladder can an 800 N worker climb before it starts to slip?



I ended up with the equation:

$d*800\sin 40 = 0.5*2000*3*sin 50$

He oneplusone! Always start these typs of mechanics question by drawing FBD. If you already have drawn it, show it to us!

 

[oneplusone]

Sorry, I don't really understand what you're saying. Can you write down the first set of equations so I can see?
thanks

EDIT: I drew a FBD, but am unable to post it here.

 

[darkxponent]

Start like this. Let the man be climbing a distance d on the ladder from the ground.

How many forces do see on the ladder?
What is the restoring Force?
At the point of slipping these forces will balance.

 

[oneplusone]

I chose the CM of the ladder as the pivot point.

I saw the frictional force (u*Normal force) and the weight of the worker

 

[darkxponent]

I chose the CM of the ladder as the pivot point.

I saw the frictional force (u*Normal force) and the weight of the worker

There are three Normal Forces on the ladder, one gravity and one Friction Force. In total there are Five Forces. Now balance the net Forces(in x and y direction separately) and net Torque. You will get three equations for three variables. Solve it.

 

[haruspex]

There are three Normal Forces on the ladder, one gravity and one Friction Force.

A 'normal' force means a force at right angles to a contacting surface. There are two contacting surfaces. Maybe you mean two normal forces, two gravitational forces, and a frictional force?
oneplusone, please post your working, not just the equation you ended up with.

 

[darkxponent]

A 'normal' force means a force at right angles to a contacting surface. There are two contacting surfaces. Maybe you mean two normal forces, two gravitational forces, and a frictional force?
oneplusone, please post your working, not just the equation you ended up with.

No i meant three Normal Forces. The person standing on the ladder will be perpendicular to the ladder surface(on one of the cyllindrical sticks). Two normal and two gravitational Force will give same answer but the Normal Forces on the ladder are three.