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A simple static equilibrium problem

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Homework Statement


Suppose we have a folding ladder, so that when its legs are spread it makes an isosceles triangle. Suppose it has a support that runs parallel to the ground. There is a bucket of paint that rests on top of the ladder, and the mass of the ladder is negligible.

Data: The legs of the ladder are 2.5m, the separation is 1.3m, and the support is 0.6m, the bucket mass a mass of 20-kg

We are supposed to find the force the support must exert to prevent the ladder from falling.

Homework Equations


Torque = r X F

The Attempt at a Solution



We are in static equilibrium so the net torque about the point where one leg hits the floor must be 0.

If you use similar triangles, you should find that the distance from the floor to the support is 1.3462m. I calculated the angle between the left leg and horizontal using trigonometry to be theta = 74.93 degrees. The angle between the vector from the floor to the support and the force of the support is 180 - theta, and the angle between the vector from the floor to the bucket and the force of gravity acting on the bucket is 90 + theta.

I set the sum equal to zero: -mg * 2.5m*sin(90 + theta) + F_support * 1.3462m * sin(180 - theta) = 0. Plug and chug and you get around 98N, but the book gets 57N.

The book has numerous examples with wrong answers, and I have personally emailed the author with agreement that these were errors. Before I jump to conclusions, I want to make sure my calculations aren't in error.


 

Answers and Replies

  • #2
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Just checking, have you drawn a diagram of the situation? This will help significantly.
 
  • #3
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Of course I have. The diagram is an isosceles triangle. That's how I calculated the distance from ground to the support along the ladder and the other calculations.
 
  • #4
haruspex
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When you say the distance from floor to support is 1.3..., is that vertically or measured up the ladder sloe?
You maybe overlooked that the weight at the top is shared between the two halves of the ladder.
 
  • #5
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When you say the distance from floor to support is 1.3..., is that vertically or measured up the ladder sloe?
You maybe overlooked that the weight at the top is shared between the two halves of the ladder.
It's measured up the ladder slope.

I considered the division of weight between the two halves, but that would still be off. I don't think you're supposed to make that assumption. I believe the weight rests and the top of the triangle so that they both receive the 20 kg.
 
  • #6
haruspex
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It's measured up the ladder slope.

I considered the division of weight between the two halves, but that would still be off. I don't think you're supposed to make that assumption. I believe the weight rests and the top of the triangle so that they both receive the 20 kg.
That makes no sense. If the weight is 20gN then by symmetry each half of the ladder supports 10gN.
I always strongly advise students to get into the habit of working entirely symbolically. Assign variables to all the given constants, only plugging in the values at the final step. There are many benefits, one of which is to make it far easier for others to follow your reasoning and spot errors.
Please repost in that form.
Assuming the shared load, I get 57.2N.
 
  • #7
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That makes no sense. If the weight is 20gN then by symmetry each half of the ladder supports 10gN.
I always strongly advise students to get into the habit of working entirely symbolically. Assign variables to all the given constants, only plugging in the values at the final step. There are many benefits, one of which is to make it far easier for others to follow your reasoning and spot errors.
Please repost in that form.
Assuming the shared load, I get 57.2N.
Using my method essentially of calculating, you didn't change anything else?
 
  • #8
haruspex
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Using my method essentially of calculating, you didn't change anything else?
I'm not sure because I did not attempt to reverse engineer the logic from the numbers. Post the algebraic steps and I'll check them.
 
  • #9
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Let d_b be the distance from the floor to the bucket along the slope which is given to be 2.5m. d_s be the distance from the floor to the support along the slope of the ladder. I found this value to be about 1.3462m by using similar triangles. Let theta be the angle between the left leg and horizontal. If you construct a right triangle by bisecting the base, you can calculate this to be 74.93 degrees. The angle vetween the vector from the floor to the support along the slope and the force of the support is 180 - theta, and the angle between the vector from the floor to the bucket along the slope and the force of gravity acting on the bucket is 90 + theta.

Torque = 0: -(1/2) m * g * d_b * sin(90 + theta) + F_support * d_s * sin(180 - theta) = 0.

Plug and chug.
 
  • #10
haruspex
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Let d_b be the distance from the floor to the bucket along the slope which is given to be 2.5m. d_s be the distance from the floor to the support along the slope of the ladder. I found this value to be about 1.3462m by using similar triangles. Let theta be the angle between the left leg and horizontal. If you construct a right triangle by bisecting the base, you can calculate this to be 74.93 degrees. The angle vetween the vector from the floor to the support along the slope and the force of the support is 180 - theta, and the angle between the vector from the floor to the bucket along the slope and the force of gravity acting on the bucket is 90 + theta.

Torque = 0: -(1/2) m * g * d_b * sin(90 + theta) + F_support * d_s * sin(180 - theta) = 0.

Plug and chug.
I see... you took moments about the point of contact with the ground, but omitted one of the forces that has a moment there. I avoided that problem by taking moments about the top of the ladder.
 
  • #11
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I see... you took moments about the point of contact with the ground, but omitted one of the forces that has a moment there. I avoided that problem by taking moments about the top of the ladder.
You're right. I missed that. I reworked it as you suggested and got 57N. Thanks.
 
  • #12
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I see... you took moments about the point of contact with the ground, but omitted one of the forces that has a moment there. I avoided that problem by taking moments about the top of the ladder.
Thanks for helping me with that problem. I'm doing self study just for the fun of it. I don't want to start another thread, but I'm confused about another problem on the same topic, and was wondering if you have any insight about it. The problem reads: A person would like to pull a car out of a ditch. This person ties one end of a chain to the car's bumper and wraps the other end around a tree so that the chain is taut. The person then pulls on the chain perpendicular to its length. If the distance between the car and tree is 5.0 m and the length of the chain between the car and the tree is 5.2 m and the person can pull with 100LB of force, what force can the chain exert on the car? The book gets roughly 180 LB. For this problem, I really have no clue other than to make some estimates. For example, I assumed the person pulling on the chain formed equilateral triangles if you draw a segment from the person straight down. Then you get the force to be 365 N so off by a lot. Any ideas?
 
  • #13
haruspex
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Thanks for helping me with that problem. I'm doing self study just for the fun of it. I don't want to start another thread, but I'm confused about another problem on the same topic, and was wondering if you have any insight about it. The problem reads: A person would like to pull a car out of a ditch. This person ties one end of a chain to the car's bumper and wraps the other end around a tree so that the chain is taut. The person then pulls on the chain perpendicular to its length. If the distance between the car and tree is 5.0 m and the length of the chain between the car and the tree is 5.2 m and the person can pull with 100LB of force, what force can the chain exert on the car? The book gets roughly 180 LB. For this problem, I really have no clue other than to make some estimates. For example, I assumed the person pulling on the chain formed equilateral triangles if you draw a segment from the person straight down. Then you get the force to be 365 N so off by a lot. Any ideas?
I think you mean isosceles triangles.
This really does need a new thread, that's a forum rule. Don't forget to post your working, such as it is. It may be that you have the wrong diagram, so post one if you can, or perhaps just describe it.

I get 182LB. I note that your answer is almost exactly double. I suspect you forget the section of chain from the person to the tree is also under tension.
 
  • #14
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I think you mean isosceles triangles.
This really does need a new thread, that's a forum rule. Don't forget to post your working, such as it is. It may be that you have the wrong diagram, so post one if you can, or perhaps just describe it.

I get 182LB. I note that your answer is almost exactly double. I suspect you forget the section of chain from the person to the tree is also under tension.
I created the thread. I can't see how I'm missing a factor of 2. You should like my thread, it's all in easily to follow algebraic symbols with the calculation at the end.
 

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