Torque and center of gravity problem

[Ben C.]

First off, here's the problem:
Suppose your lab partner has a height L of 173 cm and a weight w of 715N. You can now determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale. If the board's weight is 49N and the scale reading F is 350N, find the distance of your lab partner center of gravity from the left end of the board.

The book sets up the equation as

(350N)(173cm)-(49N)(86.5cm) = 715N* d
where d= distance from center of gravity.

I'm having trouble understanding what to use as the distance when determining torque.

1)Since the partner is resting on a board supported by two points, why is the torque provided by the partner equal to his weight * distance from center of gravity and not the total distance from the end opposite of the scale?
2) Similarly, why is the force from the scale multiplied by the height of the lab partner and not the distance from the lab partner's center of gravity?

Thanks for the help.

 

[PhanthomJay]

First off, here's the problem:
Suppose your lab partner has a height L of 173 cm and a weight w of 715N. You can now determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale. If the board's weight is 49N and the scale reading F is 350N, find the distance of your lab partner center of gravity from the left end of the board.

The book sets up the equation as

(350N)(173cm)-(49N)(86.5cm) = 715N* d
where d= distance from center of gravity.

I'm having trouble understanding what to use as the distance when determining torque.

1)Since the partner is resting on a board supported by two points, why is the torque provided by the partner equal to his weight * distance from center of gravity and not the total distance from the end opposite of the scale?
2) Similarly, why is the force from the scale multiplied by the height of the lab partner and not the distance from the lab partner's center of gravity?

Thanks for the help.

The moment (or torque) of a force about any point is equal to tjhe product of that force times the perpendicular distance from the line of action of that force to the point. The partners weight resultant acts at his cg. The board's weight resultant acts at its cg. Further, the sum of all torques of all forces about any point must equal zero, for equilibrium. It is often convenient to choose the point about which to sum forces as the point where one of the forces is unknown.

 

[kerimek]

I would first commend you for trying to understand the problem and asking for clarification. It shows that you are engaged and thinking critically about the concepts being presented to you. Now, let me address your questions:

1) The torque provided by the partner is equal to his weight multiplied by the distance from his center of gravity because torque is defined as the force applied multiplied by the distance from the pivot point. In this case, the pivot point is the center of gravity, which is the point where the entire weight of the body can be considered to act. Therefore, the torque provided by the partner is calculated using the distance from the center of gravity, not the total distance from the end opposite of the scale.

2) The force from the scale is multiplied by the height of the lab partner because the scale is supporting the entire weight of the board and the lab partner. The height of the lab partner is used because it represents the distance from the scale to the center of gravity of the system (board + lab partner). This is known as the lever arm, which is the perpendicular distance from the line of action of the force to the pivot point. In this case, the pivot point is still the center of gravity.

I hope this helps clarify your understanding of torque and center of gravity. Remember, in physics, it is important to pay attention to the definitions and concepts being used in order to correctly apply them in problem-solving. Keep up the good work!