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Triangle Ladder equilibrium problem

  • Thread starter terryds
  • Start date
  • #1
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Homework Statement



http://www.sumoware.com/images/temp/xzebnqpmgdnohmgo.png [Broken]

An isosceles-triangle-shaped ladder like the picture above, has a negligible mass. A man with mass m kg climb the ladder till the height is 3 meter (see picture above).
Determine the tension of the link (at horizontal position in the picture) between the sides of ladder.


Homework Equations


ΣF = 0
Στ = 0 (τ is moment)[/B]

The Attempt at a Solution



I've drawn the free body diagram
http://www.sumoware.com/images/temp/xzbotpidknnxeqfp.png [Broken]

I didn't draw the gravitational force of the ladder since the ladder mass is negligible.

∑Fx = 0
N4 = N3

ΣFy=0
N1+N2+Nman - Wman = 0

Then I don't know what to do.. Please help me
 
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Answers and Replies

  • #2
Since each side of the ladder is an inclined plane, some of the man's weight will exert forces parallel to the ground. Find the angle the triangle is from the ground, and use cosine to find the parallel force.
 
  • #3
ehild
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The legs of the ladder are joined by the hinge at the top. The normal forces between the hinge and the legs of equal magnitude, and of opposite direction, but not sure if they are horizontal.
The man presses the ladder, but the normal force Nman acts on the man.
You have to write the equilibrium conditions separately for both legs of the ladder.
 
  • #4
ehild
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ladder.jpg


Since each side of the ladder is an inclined plane, some of the man's weight will exert forces parallel to the ground. Find the angle the triangle is from the ground, and use cosine to find the parallel force.
The ladder is not like two inclined planes. It has steps, they are about horizontal. The man stands on one step.
 
  • #5
View attachment 78678


The ladder is not like two inclined planes. It has steps, they are about horizontal. The man stands on one step.
good point. Try solving using torque. Earth is the fulcrum, and the link counteracts the torque exerted by the man.
 
  • #6
ehild
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Try solving using torque. Earth is the fulcrum, and the link counteracts the torque exerted by the man.
I see you are new here. Welcome. If you want to help other people, try to solve the problem before giving (wrong) hints. You forget the other leg. And the torque of both the link and the weight of the man abut point A are clockwise.
 
  • #7
392
13
The legs of the ladder are joined by the hinge at the top. The normal forces between the hinge and the legs of equal magnitude, and of opposite direction, but not sure if they are horizontal.
The man presses the ladder, but the normal force Nman acts on the man.
You have to write the equilibrium conditions separately for both legs of the ladder.
Is my free body diagram correct ?
So, I can neglect the normal force between the hinge and legs, right ?
So, the FBD is
http://www.sumoware.com/images/temp/xzltagsnmkekgnnx.png [Broken]

Right ?
Or are there more forces that I can neglect ?
 
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  • #8
ehild
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The system is not a single rigid body, so you need the FBD-s separately for all pieces.
The simplest is the man: its weight is counteracted by the normal force of the ladder step.
The two legs of the ladder are separate rigid bodies, but they interact at the hinge.
Draw the FBD for both legs. What forces act on the right leg, and what forces act at the left leg?
 
  • #9
392
13
The system is not a single rigid body, so you need the FBD-s separately for all pieces.
The simplest is the man: its weight is counteracted by the normal force of the ladder step.
The two legs of the ladder are separate rigid bodies, but they interact at the hinge.
Draw the FBD for both legs. What forces act on the right leg, and what forces act at the left leg?
http://www.sumoware.com/images/temp/xzjigctgexdcnmht.png [Broken]

Hmmm..
I'm not sure this is right..
Since the left leg FBD I drew has no left direction force to balance the tension
And the right leg FBD I drew has no downward and right direction force to balance the tension and normal force.

I think there must be forces to balance those.
The contact force between hinge and the legs will balance it, right ?
So, is this the correct FBD ?
http://www.sumoware.com/images/temp/xzsqxpksqgspttgf.png [Broken]

Is it right or do I miss something ?
 
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  • #10
ehild
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It is almost right now, but remove the force Nman. It acts on the man, not on the ladder leg. The opposite to Nman acts on the ladder, but it is equal to the weight of the man. And the two upward normal forces from the ground are not equal. Use different notations.
 
  • #11
392
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So, it's gonna be like below, right ?
http://www.sumoware.com/images/temp/xzrqtoencmtbpebg.png [Broken]
 
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  • #12
ehild
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Yes, it is right now.
 
  • #13
392
13
Yes, it is right now.
I suppose that the angle Θ is the angle between floor and the leg

By the left leg,
∑Fx = 0
R sin Θ - T = 0
T = R sin Θ

∑Fy = 0
N1 + R cos Θ - W = 0
R cos Θ = W + N1

By the right leg,
ΣFx = 0
R sin Θ - T = 0
T = R sin Θ

∑Fy = 0
N2 - R cos Θ = 0
R cos Θ = N2

Then, I get
N2 = W + N1

I choose the hinge as the fulcrum, so
Torque at the left leg
∑τ = 0
N1 cos Θ . 4 - T sin Θ . 2 - W cos Θ . 1 = 0

And sin Θ is √17 / 4, cos Θ is 1/4
Till this point, is it right ?
 
  • #14
ehild
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I suppose that the angle Θ is the angle between floor and the leg
What do you mean on angle Θ? If it the angle of the force R with respect to horizontal, you can not assume it is the same the leg encloses with the floor.
Let it undetermined, with Rx and Ry components.
 
  • #15
392
13
What do you mean on angle Θ? If it the angle of the force R with respect to horizontal, you can not assume it is the same the leg encloses with the floor.
Let it undetermined, with Rx and Ry components.
Okay, so
By the left leg,
∑Fx = 0
Rx - T = 0
T = Rx

∑Fy = 0
N1 + Ry - W = 0
Ry = W + N1

By the right leg,
ΣFx = 0
Rx - T = 0
T = Rx

∑Fy = 0
N2 - Ry = 0
Ry = N2

Then, I get
N2 = W + N1

I choose the hinge as the fulcrum, so
Torque at the left leg
∑τ = 0
N1 sin Θ . 4 - T sin Θ . 2 - W cos Θ . 1 = 0
2 T sin Θ = 4 N1 sin Θ - W cos Θ
4 N1 sin Θ = 2 T sin Θ + W cos Θ
N1 = 0.5 T + (0.25 W / tan Θ)

Torque at the right leg
Στ = 0
T sin Θ . 2 - N2 sin Θ . 1 = 0
2 T sin Θ = N2 sin Θ
2 T sin Θ = (W + N1) sin Θ
2 T = mg + 0.5 T + (0.25 W / tan Θ)
1.5 T = mg + (mg / 4 tan Θ)
1.5 T = mg ( 4 tan Θ + 1) / 4 tan Θ
T = 2 mg( 4 tan Θ + 1) / 12 tan Θ
T = mg (4 tan Θ + 1) / 6 tan Θ

tan Θ is √17

So, T = mg ( 4√17 +1) / 6√17

Is it right ?
 
  • #16
ehild
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What is theta? N1 and N2 are vertical, T is horizontal. The torque is the force multiplied by the lever arm - the distance of the line of action from the fulcrum. You can find them from the given geometric data.
 
  • #17
392
13
What is theta? N1 and N2 are vertical, T is horizontal. The torque is the force multiplied by the lever arm - the distance of the line of action from the fulcrum. You can find them from the given geometric data.
Thanks.. I've got it now
 
  • #18
ehild
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Well done then! :)
 

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