Two people of different masses sit on a seesaw. M 1 , the mass of person 1, is 9

Question

Two people of different masses sit on a seesaw. M1, the mass of person 1, is 91 kg, M2 is 42 kg, d1 = 0.8 m, and d2 = 1.2 m. The mass of the board is negligible.

(a) What is the magnitude of the torque about the pivot (location A) due to the gravitational force on person 1?
|A,1| =  N · m
(b) What is the direction of this torque?

(c) What is the magnitude of the torque about location A due to the gravitational force on person 2?
|A,2| =  N · m
(d) What is the direction of this torque?

(e) Since at this instant the linear momentum of the system may be changing, we don't know the magnitude of the "normal" force exerted by the pivot. Nonetheless, it is possible to calculate the torque due to this force. What is the magnitude of the torque about location A due to the force exerted by the pivot on the board?
|A,pivot| =  N · m
(f) What is the direction of this torque?

(g) What is the magnitude of the net torque on the system (board + people)?
|A,net| =  N · m
(h) Because of this net torque, what will happen?

The seesaw will begin to rotate clockwiseThe seesaw will begin to rotate counterclockwise    The seesaw will not move.

(i) Person 2 moves to a new position, in which the magnitude of the net torque about location A is now 0, and the seesaw is balanced. What is the new value of d2 in this situation?
d2 =  m

Explanation / Answer

A)torque due to 1st man = m1gr1

= 91*9.8*0.8

= 713.44 Nm

B) INTO THE SCREEN because T = -j ×-i

C)TORQUE T2= m2gr2

= 493.92 Nm

d) direction is out of the screen becauuse T = F×r = -j×i

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