# Center of Gravity Question: Pretty Tough Conceptually

• cheechnchong
In summary, the conversation discusses a problem involving finding the correct answer to a physics question. The solution involves using the center of gravity and applying a 22N force at a given point, and summing moments about the elbow joint to solve for M. There is some confusion about which values to use and where, but ultimately the conversation provides guidance on how to correctly solve the problem.
cheechnchong
THANKS FOR THE HELP! i solved the problem successfully (correct answer too)

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The c.g. of the arm is given. Apply the 22N force there, and sum moments (torques) about the elbow joint to solve for M, noting that the sum of the moments must be 0 for equilibrium. Mind your plus and minus signs.

^^kinda vague...just want to know if my work is right?

What do i plug for W3 and it's components (of distances) seems to be the misunderstanding...does my work reflect the correct answer?

^^^Bump, i'd like to have this ASAP! I'm still trying to solve it, but I am stuck...

BUUUUUUUMP! no one helping me?

cheechnchong said:
Solving for W3 i get 1464N

I think I might be miscalculating something here...like some kinda distance--i feel that the problem is right (although conceptually i am not comfortable). if someone can point out my errors, then that'll be cool! THANKS
Does W3 mean the same thing as M? I assume it does. Pick one point (the elbow looks good to me since there is an unlabled force acting on it) and calculate the torques about that point. Be careful with your signs.

OlderDan said:
Does W3 mean the same thing as M? I assume it does. Pick one point (the elbow looks good to me since there is an unlabled force acting on it) and calculate the torques about that point. Be careful with your signs.

W3 is a force...Why do you say be careful with signs? is W3 a negative value? oh and are my measurements in the right place...can you confirm that for me?

Im checkin this thread like every few seconds lol ... so don't be surprised with quick feedback

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cheechnchong said:
W3 is a force...Why do you say be careful with signs? is W3 a negative value? oh and are my measurements in the right place...can you confirm that for me?
This is not about calculating center of gravity. Only one of the things you labeled as W is a weight. There are three forces acting on the forearm: its weight, which can be taken as acting at the CM, the muscle force acting very close to the elbow, and the normal force acting on the hand. Two of those forces are known. One is not. You need to use the torques associated with those forces to write an equation based on the fact that the arm is not rotating.

OlderDan said:
This is not about calculating center of gravity. Only one of the things you labeled as W is a weight. There are three forces acting on the forearm: its weight, which can be taken as acting at the CM, the muscle force acting very close to the elbow, and the normal force acting on the hand. Two of those forces are known. One is not. You need to use the torques associated with those forces to write an equation based on the fact that the arm is not rotating.

So the equation i am using incorrect?

cheechnchong said:
Equation Used: X(center gravity) = W1x1 + W2x2 + W3x3 / W1 + W2 + W3

My Work:
0.15 = [(111N)(0.3m) + (.15m)(22N) + W3(0.150m + 0.025m)] / (111N + 22N + W3)
Im afraid that's not the way at all. You used the definition of center of gravity, where the W1, W2 and W3 are the weights acting on the objects at positions x1, x2 and x3 respectively. The M and the force from the scale are not weights.

This is a static equilibrium problem. The pivot point (elbow joint) is not moving and nothing is rotating about it. That means the total torque about the elbow joint is zero.
There are three forces at work here. All applying a torque about the elbow joint: M (which you have to find), gravity and the scale (it's pushing upwards on the hand). The torque of a force about a point is r x F, where r is the distance from the pivot point to the point where the force is applied. The total torque must be zero. From this you can find M (watch the signs)

Galileo said:
Im afraid that's not the way at all. You used the definition of center of gravity, where the W1, W2 and W3 are the weights acting on the objects at positions x1, x2 and x3 respectively. The M and the force from the scale are not weights.

This is a static equilibrium problem. The pivot point (elbow joint) is not moving and nothing is rotating about it. That means the total torque about the elbow joint is zero.
There are three forces at work here. All applying a torque about the elbow joint: M (which you have to find), gravity and the scale (it's pushing upwards on the hand). The torque of a force about a point is r x F, where r is the distance from the pivot point to the point where the force is applied. The total torque must be zero. From this you can find M (watch the signs)

hmmmm ok this is very clear...i appreciate this--even older dan gets to this point thanks! well, i'll scratch everything out and begin anew--i'll let you know when i have trouble! again thanks

## 1. What is the definition of center of gravity?

The center of gravity is defined as the point at which the entire weight of an object can be considered to be concentrated, and where the force of gravity acts upon it.

## 2. How do you calculate the center of gravity?

The center of gravity can be calculated by finding the average of the individual centers of mass of each component of the object. This can be done by multiplying the mass of each component by its distance from a reference point, then dividing the sum of these values by the total mass of the object.

## 3. Why is the concept of center of gravity important?

The concept of center of gravity is important because it helps us understand how forces act upon objects and how they maintain their balance. It is also crucial in designing structures and machines to ensure stability and prevent tipping over.

## 4. Can the center of gravity be outside of an object?

Yes, the center of gravity can be outside of an object if there are multiple components with different masses and distances from the reference point. In this case, the center of gravity will be located at a point where the weight of the object is evenly distributed.

## 5. How does the center of gravity affect the stability of an object?

The lower the center of gravity is in an object, the more stable it will be. This is because a lower center of gravity means the weight of the object is closer to the ground, making it less likely to tip over. Additionally, objects with a larger base and lower center of gravity are more stable than those with a smaller base and higher center of gravity.

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