two students are balancing on a 10 m seesaw . the seesaw is designed so that each side of the seesaw is 5 m long. the student on the left weighs 60 kg and is sitting three meters away from from the fulcrum at the center. the student on the right weighs 45 kg. the seesaw is parallel to the ground. the Mass of the board is evenly distributed so that is center of mass is over the fulcrum. imagine that two students are sitting on the seesaw so that the torque is 0 N*m.which of the following changes will alter the torque of the seesaw?

Question

a. another student stands perfectly on the center of the seesaw.
b. both students move toward the center by one meter.
c. the heavier student moves forward 1 m, while the lighter student moves forward 1.33 m.
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Answer 1

initial balance
60 * 3 = 45 x
x = 4
so the student on the right is 4 meters from the fulcrum.

a. There is no torque resulting from more mass at the fulcrum.

b.
60 * (3-1) = 120
45 * (4-1) = 135, the student on the right hits the ground

c,
60 * (3-1) = 120
45 (4-1.33) = 120.15 pretty close to balanced.

Answer 2

So what would it be

Answer 3

Is it b

Answer 4

Yes, b