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

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Homework Help Overview

The problem involves a ladder that is 6 meters long and weighs 120 N, leaning against a smooth wall at a 50° angle with the horizontal. The question is about determining how far an 800 N worker can climb up the ladder before it begins to slip, considering the coefficient of friction between the floor is 0.5.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the angles used in torque calculations and the forces acting on the ladder. There are inquiries about the use of sine and cosine functions in the equations presented. Some participants suggest drawing a free body diagram (FBD) to clarify the forces involved.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the equations and forces at play. There is a focus on balancing forces and torques, and some guidance has been provided regarding the importance of visualizing the problem through diagrams. However, there is no explicit consensus on the approach or solution yet.

Contextual Notes

Participants note the complexity of the forces involved, including multiple normal forces and the need to balance these forces for the ladder to remain stable. There is also mention of the inability to share diagrams directly in the thread.

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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
 
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why do you have sin(40) and sin(50). would one of those not be cos?
 
But for torque, isn't it:
distance * force * sin x
 
well. if you draw the FBD the 40 degree angle is in one direction and the 50 is in the other direction.
 
oneplusone said:

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!
 
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.
 
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.
 
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
 
oneplusone said:
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.
 
  • #10
darkxponent said:
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.
 
  • #11
haruspex said:
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.
 

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