A see-saw is balanced on a pivot with two children on it. One child is sitting 1.5 m to the left of the pivot and has a mass of 50 kg. Another child of mass 30 kg is sitting on the right hand side of the pivot. What distance away from the pivot is the child on the right of the pivot?

a) 30 cm
b) 1.5 m
c) 2.5 m
d) Impossible to say without knowing the length of the see-saw

If an object is not turning, the total clockwise moment, compared to the total anti-clockwise moment about any pivot, must be what?

a) Clockwise moment is twice as large as anti-clockwise moment

b) Clockwise moment is three times as large as anti-clockwise moment

c) Clockwise moment is half as large as anti-clockwise moment

d) Clockwise moment is exactly equal in magnitude to the anti-clockwise
moment

Explanation:

Remark

This is a second class lever. It is much more efficient than the fishing pole problem. All distances are measured from the pivot in these kinds of questions.

Givens

d1 = 1.5

d2 = ?

m1 = 50 kg

m2 = 30 kg

The lighter child will have to sit further away from the pivot to make the two conditions equal.

Formula

d1*m1 = d2*m2

1.5*50 = d2 * 30

75 = 30 * d2

75/30 = d2

d2 = 2.5

Remark

Notice that the distance is longer for the lighter child. The fact that these are masses and not forces does not matter, but you should take note of it. There is a difference between masses and forces. See the fishing pole problem.

Answer to the multiple Choice question. No motion on this kind of problem means equal moments. The answer is D

Problem 2

1) The wheels are further apart making B more stable. The wider the distance the wheels are apart, the harder it would be to tip the concrete mixer over

2) The center of gravity is lower. The higher the force is the more chance you have of exerting an external force to tip the mixer over.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s