Torque seesaw problem- rotational equilibrium

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Homework Statement


A seesaw with a mass of 5 kg has one block of mass 10 kg two meters to the left of the fulcrum and another block 0.5 m to the right of the fulcrum. If the seesaw is in equilibrium,
A. find the mass of the second block.
B. find the force exerted by the fulcrum
I have trouble finding the force exerted by the fulcrum; I keep setting up the wrong equation. I do not understand why the normal forces for m1 and m2 do not come into play for this problem. Any help would be greatly appreciated.

Homework Equations


Σtorque = 0
ΣFy = 0

The Attempt at a Solution


m1 = 10 kg (2 m left of fulcrum)
m2 = 40 kg (0.5 m right of fulcrum)
seesaw mass (M) = 5 kg
F1 = upward force (normal) exerted by fulcrum onto seesaw
Mg = downward force

designated fulcrum as pivot point
up → positive
down → negative
clockwise rotation → positive
counterclockwise rotation →negative
d1 (⊥ distance from the line of action of force of m1g to pivot point) = 2 m
d2 (⊥ distance from the line of action of force of m2g to pivot point) = 0.5 m

A. ∑torque = -m1g (d1) + m2g (d2) = 0
⇒m2 = 40 kg

B. For m1,
∑Fy = n- m1g = 0
⇒ n = m1g
⇒ n = 10 kg (9.8 m/s^2)
For m2,
∑Fy = n- m2g = 0
⇒ n = m2g
⇒ n = 40 kg (9.8 m/s^2)

∑Fy = n-m1g + n-m2g - F1- Mg = 0
 

Answers and Replies

  • #2
Bystander
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See-saw? Or, hoverboard? You've omitted something.
 
  • #3
SteamKing
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I do not understand why the normal forces for m1 and m2 do not come into play for this problem. Any help would be greatly appreciated.
Why do you think this?

Homework Equations


Σtorque = 0
ΣFy = 0

The Attempt at a Solution


m1 = 10 kg (2 m left of fulcrum)
m2 = 40 kg (0.5 m right of fulcrum)
seesaw mass (M) = 5 kg
F1 = upward force (normal) exerted by fulcrum onto seesaw
Mg = downward force

designated fulcrum as pivot point
up → positive
down → negative
clockwise rotation → positive
counterclockwise rotation →negative
d1 (⊥ distance from the line of action of force of m1g to pivot point) = 2 m
d2 (⊥ distance from the line of action of force of m2g to pivot point) = 0.5 m

A. ∑torque = -m1g (d1) + m2g (d2) = 0
⇒m2 = 40 kg

B. For m1,
∑Fy = n- m1g = 0
⇒ n = m1g
⇒ n = 10 kg (9.8 m/s^2)
For m2,
∑Fy = n- m2g = 0
⇒ n = m2g
⇒ n = 40 kg (9.8 m/s^2)

∑Fy = n-m1g + n-m2g - F1- Mg = 0
The normal force is only present when considering each mass as a separate free body. When you are trying to find the reaction force at the fulcrum, the sum of the forces includes the weight of the beam and the weight of any masses resting on it.
 
  • #4
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Why do you think this?

The normal force is only present when considering each mass as a separate free body. When you are trying to find the reaction force at the fulcrum, the sum of the forces includes the weight of the beam and the weight of any masses resting on it.
So when trying to find the reactionary force (normal) exerted by the fulcrum, I should see the fulcrum and the 2 masses resting on it as a free body (free body of fulcrum) in its entirety... I mistakenly thought that adding all the upward forces and downward forces of the two masses and the fulcrum would give me the correct answer. So the upward force (normal) exerted by the fulcrum must balance the downward forces exerted by the two masses and the fulcrum in order for the seesaw to be in equilibrium. Please correct me, if I am wrong. Any help would be great. Thanks.
 
  • #5
haruspex
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I mistakenly thought that adding all the upward forces and downward forces of the two masses and the fulcrum would give me the correct answer.
Not sure what you mean by that, but you had this equation
∑Fy = n-m1g + n-m2g - F1- Mg = 0
It's a bit confusing because you reuse symbols to mean different things. If you do that others have to guess what you mean.
I think in this equation the two n's represent the normal forces on the masses from the seesaw, F1 the normal force between the fulcrum and the seesaw (though from which standpoint is unclear), and the whole is the sum of forces on... the seesaw?
If so, it is wrong to have both the n's and the m1g, m2g in there. As far as the seesaw is concerned, all that matters are the forces directly on it: the two normal forces from the masses, the normal force from the fulcrum, and gravity on itself. It can't 'feel' the gravity on the masses.
 
  • #6
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Not sure what you mean by that, but you had this equation

It's a bit confusing because you reuse symbols to mean different things. If you do that others have to guess what you mean.
I think in this equation the two n's represent the normal forces on the masses from the seesaw, F1 the normal force between the fulcrum and the seesaw (though from which standpoint is unclear), and the whole is the sum of forces on... the seesaw?
If so, it is wrong to have both the n's and the m1g, m2g in there. As far as the seesaw is concerned, all that matters are the forces directly on it: the two normal forces from the masses, the normal force from the fulcrum, and gravity on itself. It can't 'feel' the gravity on the masses.
Hi, sorry for the confusion...Yes, the equation ∑Fy = n-m1g + n-m2g - F1- Mg = 0 is the sum of the forces on the seesaw, which you mentioned is wrong. Please correct me if I am wrong, but I thought the two normal forces exerted by m1 and m2 are considered if m1 and m2 are seen as separate free bodies. Since I am looking for F1 (upward normal force) exerted by fulcrum onto seesaw, and I only consider the seesaw as a single free body, then F1, m1g, m2g, and Mg (weight exerted by seesaw) should be the only forces affecting the seesaw. Therefore, ∑Fy = +F1 - Mg - m1g - m2g = 0
 
  • #7
haruspex
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Since I am looking for F1 (upward normal force) exerted by fulcrum onto seesaw, and I only consider the seesaw as a single free body, then F1, m1g, m2g, and Mg (weight exerted by seesaw) should be the only forces affecting the seesaw. Therefore, ∑Fy = +F1 - Mg - m1g - m2g = 0
Yes, except that, strictly speaking, if you are considering the seesaw as the free body then you should write N1, N2 for normal forces from the masses. In separate equations for those masses as free bodies you then obtain N1=m1g etc.
If you take the seesaw together with the two masses as your (rigid) body then you can just sum the masses and get the equation above directly.
 
  • #8
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Yes, except that, strictly speaking, if you are considering the seesaw as the free body then you should write N1, N2 for normal forces from the masses. In separate equations for those masses as free bodies you then obtain N1=m1g etc.
If you take the seesaw together with the two masses as your (rigid) body then you can just sum the masses and get the equation above directly.
Thank you so much! This clarified my questions about this problem.
 

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