Are My Calculations on Ladder Equilibrium and Torque Correct?

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The discussion centers on the equilibrium conditions of a uniform ladder measuring 6.0m in length and weighing 12.0kg, positioned against a smooth vertical wall. The calculated normal force from the wall (N1) is 117.72 Newtons, while the force exerted by the ground (N2) is determined to be 33.98 Newtons. The minimum coefficient of friction required for stability is calculated to be 0.289. The calculations presented are accurate based on the principles of static equilibrium and torque analysis.

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buffgilville
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1) Give the conditions for the equilibrium of an extended object.
An extended object is at equilibrium when no force is acting on it.

2) A uniform ladder of length 6.0m and mass 12.0kg is placed against a smooth vertical wall. The ground is rough.

a) Calculate the force on the ladder from the wall.
vertical forces = 0 = N1 - W
so N1 = 117.72
horizontal forces = 0 = F - N2
Total Torque = 0
0 = F(0) + N1(0) + (12*9.81*cos60*(6/2)) - (N2sin60(6))
so, N2 = 33.98 Newtons

b) Find the minimum coefficient of friction of the ground.
F - 33.98 = 0 ---> F=33.98
friction > F1/N1
so, coefficient of friction=33.98/117.72 ----> 0.289

Are my answers correct? If not, what did I do wrong. Please help. Thanks! :smile:
 
Last edited:
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buffgilville said:
1) Give the conditions for the equilibrium of an extended object.
An extended object is at equilibrium when no force is acting on it.

2) A uniform ladder of length 6.0m and mass 12.0kg is placed against a smooth vertical wall. The ground is rough.

a) Calculate the force on the ladder from the wall.
vertical forces = 0 = N1 - W
so N1 = 117.72
horizontal forces = 0 = F - N2
Total Torque = 0
0 = F(0) + N1(0) + (12*9.81*cos60*(6/2)) - (N2sin60(6))
so, N2 = 33.98 Newtons

b) Find the minimum coefficient of friction of the ground.
F - 33.98 = 0 ---> F=33.98
friction > F1/N1
so, coefficient of friction=33.98/117.72 ----> 0.289

Are my answers correct? If not, what did I do wrong. Please help. Thanks! :smile:

I'm not going over your problem,but my guess is that:
An extended object is at equilibrium when the resulting momentum of all forces acting on it is nil and the result of all forces acting on it is nil.
It's placing a zero at the left of the dynamics' second law for the rotation movement and for the translation one.
 
Last edited:
What's nil ?
 
buffgilville said:
What's nil ?

ZERO,NAUGHT,NOTHING,ZIP,LOVE ?
Don't u like football??
 
Can someone help me check my work on the other problems, and tell me if they are correct?
 

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