Question about static equilibrium

Click For Summary

Homework Help Overview

The discussion revolves around a static equilibrium problem involving a ladder resting against a wall. The ladder's weight, angle, and the forces acting on it are central to the inquiry. The original poster is trying to determine the force required to make the ladder slide at the bottom while considering the effects of static friction and normal forces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the forces and moments acting on the ladder, expressing uncertainty about the relationship between the force of pull and the conditions for sliding versus tipping. Some participants question the completeness of the force equations and suggest the need for a free body diagram. Others raise the issue of multiple unknowns and the necessity of balancing forces and torques.

Discussion Status

Participants are actively engaging with the problem, offering insights into the equilibrium conditions and the relationships between forces. There is an ongoing exploration of the equations needed to describe the system, with some guidance provided on how to approach the problem symbolically. However, no consensus has been reached on the correct interpretation or solution.

Contextual Notes

Participants note the complexity of the problem due to the presence of multiple forces and the requirement to balance them in a two-dimensional static scenario. There is also mention of the need to clarify the direction of the applied force and the role of static friction in the context of the problem.

bioengi12345
Messages
5
Reaction score
0

Homework Statement

A Ladder weighs 20 Lbs. IT rests against a smooth wall at point B. Theta=45. If you pull the ladder at its midpoint, how much force is needed to make it slide at point A? u=.6.
The ladder is 15 ft long, Point b is the top and point A is the bottom. 45 Deg is the bottom angle.
Force of pull= 20lb. Now i think the number worked out this was just cause the angle is 45 degrees and gravity as well as the force of the pull act about the same point, however, this just seems like that it would take to tip the ladder backwards, not make it slide. I am not really sure where to go from here.

Homework Equations


(sum of forces in x dir) : Force of pull - force of static friction = 0

(sum of forces in y dir) : -20 + normal force at A = 0

(sum of moments about A) : -20lb(5.3m)-force of pull(5.3m) = 0

The Attempt at a Solution


Force of pull= 20lb. Now i think the number worked out this was just cause the angle is 45 degrees and gravity as well as the force of the pull act about the same point, however, this just seems like that it would take to tip the ladder backwards, not make it slide. I am not really sure where to go from here.
 
Physics news on Phys.org
Anyone have any ideas?
 
You forgot the normal force from the wall.
To make the ladder just slide, but not tip over, both the resultant force components and the torque have to be zero.
 
So it would be: Force of pull - Force of static friction - normal force of the wall=0? Then i have 3 unknowns. Sorry I am normally really good at this stuff, the whole idea of equilibrium gets me sometimes.
 
bioengi12345 said:
So it would be: Force of pull - Force of static friction - normal force of the wall=0? Then i have 3 unknowns. Sorry I am normally really good at this stuff, the whole idea of equilibrium gets me sometimes.
Three unknowns is fine since in 2D statics problems you have three equations.
Is the direction of pull given?
 
pulling horizontally away from the wall at the midpoint.
 
This problem actually has four unknowns, but also four equations. (although only three are needed to solve it)
 
bioengi12345 said:
pulling horizontally away from the wall at the midpoint.
Ok, so have you drawn a free body diagram? Why do you write "force of pull = 20 lbs"? Isn't that what we're to find?
Create symbols for the known and unknown forces and lengths. Work symbolically until the final step. Write three equations: one for horizontal force balance, one for vertical force balance, and one for torque balance. (Choose a reference axis for the torque.)
 
Im trying to say that is what i got as the answer but i don't think its right. I haven't made a whole lot of progress here some direction would be very helpful. i have set up 3 equations. sum of f in x ( fp-fs-nB=0, -20+nA=0, -20(5.3m)+fP(5.3)+nB(10.6)=0 ... is any of this looking right? i need to get this turned in so some solid hint would be very nice.
 
  • #10
bioengi12345 said:
Im trying to say that is what i got as the answer but i don't think its right. I haven't made a whole lot of progress here some direction would be very helpful. i have set up 3 equations. sum of f in x ( fp-fs-nB=0, -20+nA=0, -20(5.3m)+fP(5.3)+nB(10.6)=0 ... is any of this looking right? i need to get this turned in so some solid hint would be very nice.
When you put all of that together, what do you get for fP ?
 
  • #11
bioengi12345 said:
Im trying to say that is what i got as the answer but i don't think its right. I haven't made a whole lot of progress here some direction would be very helpful. i have set up 3 equations. sum of f in x ( fp-fs-nB=0, -20+nA=0, -20(5.3m)+fP(5.3)+nB(10.6)=0 ... is any of this looking right? i need to get this turned in so some solid hint would be very nice.
First, you had not previously posted that version of your equations; second, you have a sign wrong in the x direction balance; third, you would do well to work entirely symbolically as I suggested (it avoids the units conversion); fourth, you have not used the fact that you are interested in the case where the friction is only just enough to balance the forces.
I presume you no long claim 20lbs as an answer.
 
  • #12
bioengi12345 said:
Im trying to say that is what i got as the answer but i don't think its right. I haven't made a whole lot of progress here some direction would be very helpful. i have set up 3 equations. sum of f in x ( fp-fs-nB=0,
Check the sign in front of nB.
Use the relation between NA and fs.
 

Similar threads

Static equilibrium | Uniform sphere on incline | Determining friction
  • · Replies 43 ·
2
Replies
43
Views
3K
Is there any friction in this case?
  • · Replies 12 ·
Replies
12
Views
1K
Static equilibrium -- interpretation of forces
  • · Replies 10 ·
Replies
10
Views
3K
How Do You Calculate Torques for a Ladder in Static Equilibrium?
  • · Replies 2 ·
Replies
2
Views
2K
A simple static equilibrium problem
  • · Replies 13 ·
Replies
13
Views
3K
Direction of force exerted by ground for the ladder problem
  • · Replies 2 ·
Replies
2
Views
4K
Why Isn't the Sum of Torques Zero in This Ladder Equilibrium Problem?
  • · Replies 8 ·
Replies
8
Views
7K
Trouble with Static Equilibrium
  • · Replies 5 ·
Replies
5
Views
2K
Mass of a picture in static equilibrium
  • · Replies 8 ·
Replies
8
Views
2K
Static equilibrium - man on a ladder
  • · Replies 6 ·
Replies
6
Views
6K