How does the math for lever forces work in this scenario?

  • Thread starter Thread starter jackrabbit
  • Start date Start date
  • Tags Tags
    Forces Lever
AI Thread Summary
The discussion centers on understanding the mechanics of lever forces in a scenario involving a horizontal rod subjected to various forces. The upward force of 11 pounds at the center and downward forces of 2 pounds and 8 pounds create a complex interaction around Point X, which is incorrectly assumed to be the fulcrum. The correct analysis shows that the moments created by these forces result in a net clockwise moment about Point X, leading to the rod dipping down on the side with the 8-pound force. The calculations reveal that the net moment is 15 clockwise, confirming the rod's tilt. The conversation emphasizes the importance of accurately identifying the fulcrum and calculating moments to understand lever mechanics.
jackrabbit
Messages
11
Reaction score
0
Point A
Point X 11 pounds
5 inches 5 inches ↑ 5 inches 5 inches
_________________________________________________________________
2 pounds 8 pounds
↓ ↓


Assume the horizontal line is a rod that is 20 inches long. The 11 pounds upward force is from a cable tied to the middle of the rod at Point A. There is a 2 pound downward force at left end (10 inches to the left of center), and an 8 pound downward force at 5 inches to the right of center. Point X is five inches to the left of center and is just a fixed point (imagine it is a rod sticking out, and that the horizontal rod runs into when it is lifted).

I am trying to figure out if the horizontal rod will rotate around point X, using the basic formula for force re a lever (i.e., the total force at the fulcrum is equal to force on the lever multiplied by distance to the fulcrum).

I experimented with this and found out that the horizontal rod dipped down on the end with 8 pounds. I was trying to understand the math.

My hypothesis, which was completely wrong, was as follows:

First, I thought of Point A as a fulcrum. Which would mean that the net force at this point was the difference between 8 x 5 and 2 x 10, i.e., a net downward force of 20

Then, turning to Point X, I thought the net force would be the difference between 11 x 5 (upward force) and 20 x 5 (downward force), which would mean a net upward force of 15. This would mean the rod would tilt up on the right side.

So, that was obviously wrong, and probably laughably so to those who actually understand this.

Can you please help me? How does the math here work?
 
Physics news on Phys.org
jackrabbit said:
Point A
Point X 11 pounds
5 inches 5 inches ↑ 5 inches 5 inches
_________________________________________________________________
2 pounds 8 pounds
↓ ↓


Assume the horizontal line is a rod that is 20 inches long. The 11 pounds upward force is from a cable tied to the middle of the rod at Point A. There is a 2 pound downward force at left end (10 inches to the left of center), and an 8 pound downward force at 5 inches to the right of center. Point X is five inches to the left of center and is just a fixed point (imagine it is a rod sticking out, and that the horizontal rod runs into when it is lifted).

I am trying to figure out if the horizontal rod will rotate around point X, using the basic formula for force re a lever (i.e., the total force at the fulcrum is equal to force on the lever multiplied by distance to the fulcrum).
This is an incorrect statement. The total moment (or torque) about the pivot is equal to the sum of the individual moments of the forces about the pivot. A clockwise moment may be considered as a negative moment, and a counterclockwise moment is then considered to be positive.
I experimented with this and found out that the horizontal rod dipped down on the end with 8 pounds. I was trying to understand the math.

My hypothesis, which was completely wrong, was as follows:

First, I thought of Point A as a fulcrum. Which would mean that the net force at this point was the difference between 8 x 5 and 2 x 10, i.e., a net downward force of 20
point A is not the fulcrum
Then, turning to Point X, I thought the net force
that should be net moment
would be the difference between 11 x 5 (upward force) and 20 x 5 (downward force), which would mean a net upward force of 15. This would mean the rod would tilt up on the right side.

So, that was obviously wrong, and probably laughably so to those who actually understand this.

Can you please help me? How does the math here work?
About the fulcrum point X, The 2 pound force produces a moment of 2(5) = 10, counterclockwise (ccw); the 11 pound force produces a moment of (11)(5) = 55 ccw; the 8 pound force produces a moment of (8)(10) = 80 clockwise (cw). So the net moment about the fulcrum is __________?? and it is cw or ccw? And I'm not laughing. :wink:
 
Thanks. This is great.

So, 15 cw and the rod points down on the right side.

I had considered that possibility at one point but dismissed it since I thought point X could not be the fulcrum with respect to the 2 pound force since it is actually lifting the rod down off of point X, rather acting on point X directly. The only "fulcrum" (which I see now is not a fulcrum) it is acting directly on is point A. Well, that's what I get for making things overly complicated.

Thanks again for the clear and quick answer. Much appreciated.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top