Question

1. Person A has twice the mass of person B. If they are placed at both sides of the seesaw and are at equilibrium, then

A. person a is half as far from the folcrum as person B

B. person A is 1/4 as far from the folcrum as person B

C. person A is 4 times farther away from the folcrum as person B

D. person A is twice as far away from the folcrum as person B

Answer 1

For the seesaw to be in equilibrium, the sum of moments about a point must be equal for there to be in equilibrium. Moment of force is the force multiplied by the distance taken from the fulcrum i.e

F_A*x_A = F_B*x_B ............................................ equation 1

where F_A  and F_B are the forces of person A and B respectively

and x_A and x_B are the distances from the fulcrum of person A and B respectively.

Since the total distance of seesaw is not given we will do the back calculation, taking the options one by one and check the equilibrium condition.

From equation 1 we have

m_A*g*x_A = m_B*g*x_B where g is the gravitational constant.

or m_A*x_A = m_B*x_B ..................................... equation 2

Given: m_A = 2*m_B

Taking the first option

a) x_A = 1/2*x_B

Putting values in equation 2 we get

2*m_B*1/2*x_B = m_B*x_B

or m_B*x_B = m_B*x_B

since LHS is equal to RHS we can say that the equation is satisfied. This means that the person A is half the distance from the fulcrum as person B.

b) Checking the second option which says that

x_A = 1/4*x_B

Using equation we get

2*m_B*1/4*x_B = m_B*x_B

or 1/2*m_B*x_B = m_B*x_B

Since the LHS is not equal to RHS. It shows that the fulcrum is not in equilibrium.

Likewise, we can check the other options too.

So, the correct option is a).