Static equilibrium & force of friction

In summary: The first term is the normal reaction force of the wall, with \ell being the length of the ladder. The second term is the weight if the ladder and the final term is the weight of the man.Do you follow? If so, can you do the same for...T=FN(10)sin65-22(2.8)x1/2cos65=0T=FNr=(95)(9.8)r
  • #1
janelle1905
25
0

Homework Statement


A window cleaner of mass 95 kg places a 22-kg ladder against a frictionless wall at an angle of 65 degrees with the horizontal. The ladder is 10m long and rests on a wet floor with a coefficient of static friction equal to 0.40. What is the maximum length that the window cleaner can climb before the ladder slips?


Homework Equations


T = rF


The Attempt at a Solution


Sin65 = opp/10, therefore opposite = 9.06
Cos65 = adj/10, therefore adjacent = 4.23
m = 22; M=95
Max Ffr = uFN = u(mg + Mg) = Fw
Fy=FN-mg-Mg
T=Fw(9.06)-mg(4.23/2)

Any help is appreciated!
 
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  • #2
How do you know that the normal force (at the base of the ladder) will be equal to the combined weight of the ladder and man?
 
  • #3
Hootenanny said:
How do you know that the normal force (at the base of the ladder) will be equal to the combined weight of the ladder and man?

I originally thought it would be equal to that because (mg + Mg) is the force exerted onto the ground ... but I see that my logic there is wrong.
Should it be an unknown?
Sorry, I am very lost with this question!
 
  • #4
janelle1905 said:
I originally thought it would be equal to that because (mg + Mg) is the force exerted onto the ground ... but I see that my logic there is wrong.
Should it be an unknown?
Sorry, I am very lost with this question!
I would start by examining the turning moments about the top of the ladder, i.e. assume that the point where the ladder touches the wall is the pivot.
 
  • #5
By turning moments, do you mean torque?
I thought this would be the correct eq'n for torque:
T=Fw(9.06)-mg(4.23/2)

And I know that T must equal 0 for the system to be at equilibrium.
 
  • #6
janelle1905 said:
By turning moments, do you mean torque?
I thought this would be the correct eq'n for torque:
T=Fw(9.06)-mg(4.23/2)
Yes, I mean torques. What do each of those terms represent?
janelle1905 said:
And I know that T must equal 0 for the system to be at equilibrium.
Indeed, that is correct.
 
  • #7
Hootenanny said:
Yes, I mean torques. What do each of those terms represent?

Because T=rF
T = Fw(9.06) - mg(4.23/2)
The first force is Fw (= force of wall)
9.06 is the height of the wall (i.e. where the ladder touches the wall)

The negative force, is given by F=mg
And r=4.23/2

I'm sorry that I am so confused by this question, but I appreciate your help a lot!
 
  • #8
janelle1905 said:
Because T=rF
T = Fw(9.06) - mg(4.23/2)
The first force is Fw (= force of wall)
9.06 is the height of the wall (i.e. where the ladder touches the wall)

The negative force, is given by F=mg
And r=4.23/2
Ahh, I see, you are taking moments about the base of the ladder, rather than the top. No matter, since you are going to have to do both eventually. However, I think that you're forgetting one small thing ...
... what about the man?
janelle1905 said:
I'm sorry that I am so confused by this question, but I appreciate your help a lot!
No need to apologise, we wouldn't be here if students weren't asking for help! :smile:
 
  • #9
Okay - so does the torque equation have to be re-written to include the mass of the man? And is that then the unknown?
 
  • #10
janelle1905 said:
Okay - so does the torque equation have to be re-written to include the mass of the man?
Indeed it does.
janelle1905 said:
And is that then the unknown?
Yes, that will be an unknown, but your also going to have another unknown: the reaction force from the wall.
 
  • #11
Okay...thanks for your help...and sorry for my delay in responding, but I still haven't had much success with this question.
I have the torque equation as:
T=FN(10)sin65-22(2.8)x1/2cos65=0
as well as: T=FNr=(95)(9.8)r

But I don't know how to incorporate the mass of the man into the normal force of the ladder (to calculate force of static friction).
 
  • #12
janelle1905 said:
Okay...thanks for your help...and sorry for my delay in responding, but I still haven't had much success with this question.
I have the torque equation as:
T=FN(10)sin65-22(2.8)x1/2cos65=0
as well as: T=FNr=(95)(9.8)r

But I don't know how to incorporate the mass of the man into the normal force of the ladder (to calculate force of static friction).
I'm not quite sure what your doing there, your torque equation should have three terms in it: one term for the weight of the ladder, one term of the weight of the man and one term for the normal reaction force of the ladder.

Try drawing a diagram of the situation. I'll help you out and write the torque equation if we take moments about the base of the ladder (i.e. we take the base of the ladder as the pivot).

[tex]F_N\ell\sin\theta - W_\ell\frac{\ell}{2}\cos\theta - W_mx\cos\theta = 0[/tex]

The first term is the normal reaction force of the wall, with [itex]\ell[/itex] being the length of the ladder. The second term is the weight if the ladder and the final term is the weight of the man.

Do you follow? If so, can you do the same for the torques about the upper point of the ladder?
 
Last edited:

1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and has no net force acting on it. This means that all forces acting on the object are balanced and cancel each other out.

2. How is static equilibrium different from dynamic equilibrium?

Static equilibrium is when an object is at rest, whereas dynamic equilibrium is when an object is in motion at a constant speed. In dynamic equilibrium, the net force on an object is still zero, but the object is moving.

3. What role does the force of friction play in static equilibrium?

The force of friction is an important factor in static equilibrium as it helps to keep an object from moving when a force is applied to it. Friction acts in the opposite direction of the applied force, helping to balance out the forces and keep the object at rest.

4. How does the force of friction change with different surfaces?

The force of friction depends on the type of surface the object is on. Rougher surfaces tend to have a greater force of friction, while smoother surfaces have a lower force of friction. Additionally, the force of friction also depends on the weight of the object and the angle at which the force is applied.

5. How can the force of friction be calculated?

The force of friction can be calculated using the equation Ff = μN, where Ff is the force of friction, μ is the coefficient of friction, and N is the normal force. The coefficient of friction is a constant that depends on the type of surfaces in contact, while the normal force is the force exerted by the surface on the object in a direction perpendicular to the surface.

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