Solving Weight Distribution for Iron Bar: 2 Man Problem

In summary: That man's force and the weight of the bar are equal and opposite. So the total weight that man is supporting is the weight of the bar plus the weight of the other man.In summary, for the first part, an iron bar of weight W is supported by a man and the ground, with the man supporting half of the weight. For the second part, if the bar is balanced by two men, the weight experienced by one of the men at the instant the other man lets go is the weight of the bar plus the weight of the other man. This problem involves moments and torque, and can be solved using Newton's second law and principles of linear and angular acceleration.
  • #1
konichiwa2x
81
0
An iron bar of weight W is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle alpha with the horizontal. What is the weight experienced by the man?

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?


Can someone please give me an idea of how to do these two problems? thanks for your time.
 
Last edited:
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  • #2
Think moments.
 
  • #3
This question was under the chapter "Laws of motion". Principle of moments hasnt been taught to us yet.
 
  • #4
konichiwa2x said:
This question was under the chapter "Laws of motion". Principle of moments hasnt been taught to us yet.
Does "Laws of motion" include angular motion? Perhaps you have encountered the term torque rather than the term moment.
 
  • #5
I know torque = Force x distance
also torque = moment of intertia . angular acceleration.

but I still don't understand how to do this problem using this.
 
  • #6
konichiwa2x said:
I know torque = Force x distance
also torque = moment of intertia . angular acceleration.

but I still don't understand how to do this problem using this.
For the first part, the bar is at rest. The net force acting on the bar must be zero and the sum of the torques must be zero. You treat the bar as if all the weight was acting at its center of gravity (the middle of the bar). You can calculate the torques about any point you want. By setting the total force to zero and the torque to zero you will get two equations for the two things you do not know, which are how much weight the ground is supporting and how much weight the man is supporting. That should get you started.
 
  • #7
ok here goes

let F1 and F2 be the forces on the bar due to the ground and the man respectively. Let B be the centre of the bar. A and C are the points on the ground and on the man respectively.

so W = F1 + F2
Net torque must be 0.
F1 x AB = F2 x BC
F1.AB.sin(a) = F2.BC.sin(a)
F1=F2 (AB = BC)

so W = 2F1=2F2

F2 = F1 = W/2

Is this correct?

I am sorry.. This must be irritating but I still don't understand how to do this part of the problem :

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?

thanks for your time.
 
Last edited:
  • #8
konichiwa2x said:
ok here goes

let F1 and F2 be the forces on the bar due to the ground and the man respectively. Let B be the centre of the bar. A and C are the points on the ground and on the man respectively.

so W = F1 + F2
Net torque must be 0.
F1 x AB = F2 x BC
F1.AB.sin(a) = F2.BC.sin(a)
F1=F2 (AB = BC)

so W = 2F1=2F2

F2 = F1 = W/2

Is this correct?

I am sorry.. This must be irritating but I still don't understand how to do this part of the problem :

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?

thanks for your time.
First part looks good. For the second part, whatever force that man is exerting does not contribute to the torque about his point of contact. The torque that is present will cause the bar to rotate, accelerating its center. At the first instant, this is a linear acceleration downwards. Newton's second law applies.
 

Related to Solving Weight Distribution for Iron Bar: 2 Man Problem

1. How do you determine the weight distribution for an iron bar in a 2-man problem?

The weight distribution for an iron bar in a 2-man problem can be determined by using the principle of moments. This involves balancing the moments (force x distance) on both sides of the bar to ensure it remains in equilibrium. The weight distribution can be calculated by dividing the total weight of the bar by the distance from each end to the center of mass.

2. What factors affect the weight distribution in a 2-man problem?

The weight distribution in a 2-man problem can be affected by various factors such as the length and weight of the bar, the distance between the two men, and the positioning of the men on the bar. These factors can influence the center of mass and therefore impact the weight distribution.

3. How does the positioning of the men on the bar affect the weight distribution?

The positioning of the men on the bar is crucial in determining the weight distribution. If the men are not positioned symmetrically on either side of the center of mass, it can cause an imbalance and affect the weight distribution. It is important to ensure that the distance from each man to the center of mass is equal.

4. What is the importance of solving weight distribution in a 2-man problem?

Solving weight distribution in a 2-man problem is important to ensure that the bar remains in equilibrium and does not tip over. It also helps in determining the amount of weight each man must support and can prevent any potential injuries or accidents.

5. Are there any other methods for solving weight distribution in a 2-man problem?

Yes, there are other methods for solving weight distribution in a 2-man problem such as using the law of moments, which states that the sum of the clockwise moments must be equal to the sum of the counterclockwise moments for the bar to remain in equilibrium. Additionally, computer simulations and mathematical equations can also be used to determine the weight distribution.

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