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.