Easy beginner torque/com question. please

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Homework Help Overview

The problem involves a uniform ladder leaning against a wall, requiring the calculation of the minimum friction needed to prevent slipping. The context is torque and equilibrium in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the ladder, including gravitational force, normal forces, and friction. There are attempts to establish equilibrium conditions and torque calculations. Some express confusion about the free body diagram and the forces involved.

Discussion Status

The discussion includes various attempts to clarify the forces and torques acting on the ladder. Some participants have provided guidance on writing equilibrium equations, while others continue to express uncertainty about their understanding of the problem setup.

Contextual Notes

Participants mention difficulties with the free body diagram and the relationships between different forces. There is a reference to a specific angle and the need for detailed work to clarify the situation.

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Homework Statement



A uniform 15kg ladder whose length is 5.0 m stands on the floor and leans against a vertical wall,
making an angle of 25 with the vertical. Assuming that the friction between the ladder and the wall
is negligible, what is the minimum amount of friction between the ladder and the floor which will
keep the ladder from slipping.
(34 N)

Homework Equations

torque = fperp * distance

The Attempt at a Solution



I keep on writing stuff but never get the right answer (which is 1/2 tan theta * mg)
i have no idea how to get there. i keep mapping and drawing FBD and i can't get there... i think its my understanding of torque and not understanding where the pivot point on this ladder is...

anyone care to share
 
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Can you show your work in detail?

ehild
 
ehild said:
Can you show your work in detail?

ehild

sure 1 sec let me write neatly
 
okay past this point I am not sure what to do..i assume the force wanting it to slip is Force-> (the horizontal ocmponent of MG) so i try to equate Ff (Friction) to The Force -> (Fparallelsintheta) but i get the answer 68 not 34...

http://i44.tinypic.com/2vmvalg.jpg

i realize my problem lays in the fact i don't know what forces are acting on what and my free body diagram is probably wrong
 
Last edited:
The ladder is in rest, so the resultant force is zero and the torque is also zero.

You have vertical forces acting on the ladder: mg and FN (from the ground) and horizontal ones: FWN (the normal force from the wall) and the force of static friction, Ff.

How are the magnitudes of these forces related?

The forces have torque. Choose an appropriate point to write it up.

ehild
 
ehild said:
The ladder is in rest, so the resultant force is zero and the torque is also zero.

You have vertical forces acting on the ladder: mg and FN (from the ground) and horizontal ones: FWN (the normal force from the wall) and the force of static friction, Ff.

How are the magnitudes of these forces related?

The forces have torque. Choose an appropriate point to write it up.

ehild

this doesn't help me =[
 
You do not need Fparallel. Write up the equilibrium for both the vertical and horizontal forces separately.
 
Last edited:
ehild said:
You do not need Fparallel. Write up the equilibrium for both the vertical and horizontal forces separately.

my problem is i don't know what forces are where
 
martinlematre said:
my problem is i don't know what forces are where

See figure. It is identical to your one.

mg acts at the midpoint of the ladder, vertically down. FWN acts at the upper end of the ladder, horizontal, away from wall. N and Ff act at the lower end of the ladder, N vertically up, Ff horizontally, towards the wall, as it is against slipping the ladder outward.
 

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  • #10
2 hr later i am still stuck ty..
 
  • #11
The ladder is a rigid body. To be in equilibrium (no motion) the sum of all forces has to be zero both the horizontal and vertical components. Also the torque has to be zero.

Do you see the forces? The ladder can not move vertically, so what forces have to cancel?
It can not move horizontally, so the horizontal forces have to cancel.
It can not rotate about its upper end so the sum of torques is zero.

Can you write up the corresponding equations?

ehild
 
  • #12
ehild said:
The ladder is a rigid body. To be in equilibrium (no motion) the sum of all forces has to be zero both the horizontal and vertical components. Also the torque has to be zero.

Do you see the forces? The ladder can not move vertically, so what forces have to cancel?
It can not move horizontally, so the horizontal forces have to cancel.
It can not rotate about its upper end so the sum of torques is zero.

Can you write up the corresponding equations?

ehild
i got it no biggie
 
  • #13
So you have solved the problem?

ehild
 
  • #14
ehild said:
So you have solved the problem?

ehild

No I've just decided I am going to go into liberal arts instead
 

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