Verification of a ladder aginst a wall force

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The discussion revolves around calculating the force exerted by a wall on a ladder in equilibrium, with a painter climbing it. The original poster believes their answer of 320 N differs from the teacher's 280 N due to the consideration of torques. Participants debate the importance of both torque and force equilibrium, with some emphasizing that the problem is primarily about rotational equilibrium. There is consensus that the frictional force at the base of the ladder is necessary for stability, but its exact value is not critical to solving the problem. The conversation highlights the complexities of statics problems, particularly when friction is not explicitly mentioned.
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Homework Statement


A 20 kg ladder rests against a smooth wall as shown. If a painter (50 kg) climbs 4.0 meters up a 6 meter long ladder, find the force, F1, the wall exerts on the ladder. The angle between the ladder and the ground is 53 deg.

Homework Equations


CW torque must = CCW torque

The Attempt at a Solution


The teacher posted an answer of 280 but I get 320. I think I have solved the problem correctly.

See my attached diagram
 

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barryj said:

Homework Statement


A 20 kg ladder rests against a smooth wall as shown. If a painter (50 kg) climbs 4.0 meters up a 6 meter long ladder, find the force, F1, the wall exerts on the ladder. The angle between the ladder and the ground is 53 deg.

Homework Equations


CW torque must = CCW torque

The Attempt at a Solution


The teacher posted an answer of 280 but I get 320. I think I have solved the problem correctly.

See my attached diagram
For ladder to be in equilibrium there is net force and torque will be zero how can you justify that there is net force is zero?
 
If you assume that the ladder pivots at the bottom, I think only the torques from the ladder weight, the painter weight, and the force on the ladder at the top matter.
 
barryj said:
If you assume that the ladder pivots at the bottom, I think only the torques from the ladder weight, the painter weight, and the force on the ladder at the top matter.
Yes this is for net torque ok but what about net force for transportation equilibrium? Look again force is balacing?
 
Yes, there is a force applied to the bottom of the ladder but this force will not affect the torque. The only forces that affect the torque are the wall force, the ladder weigh, and the painter, yes??
 
barryj said:
Yes, there is a force applied to the bottom of the ladder but this force will not affect the torque. The only forces that affect the torque are the wall force, the ladder weigh, and the painter, yes??
Why you are thinking about only torque for translation equilibrium ΣFx=0,ΣFy=0 are these conditions satisfying?
 
I am not concerned with translation equilibrium, only rotational equilibrium. I know that when I sum Fy forces that the Fy force at the bottom of the ladder is equal to the weight of the ladder + the weight of the painter = 686 N but so what? The wall force will be equal to the friction force at the bottom of the ladder but again , so what. If I assume the ladder pivots about the bottom, I only need to consider the torques from the wall force, the painter, and the ladder weight as the forces at the bottom do not enter into the torque equation.
 
barryj said:
I get 320.
So do I.
Abhishek kumar said:
Why you are thinking about only torque
Because the normal and frictional forces from the ground will satisfy the linear equations, but we do not need to find them.
 
The problem actually has two parts. The first part is to determine the force on the ladder from the wall. The second part is to determine the coefficient of friction between the ladder and the ground. It seems to me that the Fnormal will be just the addition of the weight of the ladder and the painter as the Force from the wall does not enter into the force normal. Then the Fx would be the force from the wall. So the friction force (horizontal) is 320 N and the normal force is 686 N. However, this does not allow the calculation of the coefficient of friction. Just knowing what the friction is does not allow the calculation of the cooef of friction.
 
  • #10
haruspex said:
So do I.

Because the normal and frictional forces from the ground will satisfy the linear equations, but we do not need to find them.
I know that.In question he had not mentioned frictional force that's why i asked.
 
  • #11
Abhishek kumar said:
I know that.In question he had not mentioned frictional force that's why i asked.
It is posed as a statics question, and the wall is smooth, so there has to be adequate friction from the ground, but it is of no interest. The OP method was clearly appropriate.
 
  • #12
haruspex said:
It is posed as a statics question, and the wall is smooth, so there has to be adequate friction from the ground, but it is of no interest. The OP method was clearly appropriate.
But you have to mention that there is friction between ground and ladder
 
  • #13
Abhishek kumar said:
But you have to mention that there is friction between ground and ladder
If the ladder can slip the question makes no sense. The failure to mention friction might irritate a perfectionist, but quibbling over that is of no help in answering the posted query.
 
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