Two people (from the 1950s!) sit on opposite sides of a uniform board of length L. One person, of mass m1 = 50.0 kg, sits at a point 1.90 m to the left of the fulcrum. The fulcrum is located at the center of the board. The other person, of mass 45.0 kg, sits at a point a distance x2 to the right of the fulcrum.
Calculate the value of x2 required for the system to be in rotational equilibrium. Make sure to enter units along with your numerical value, which should have 3 significant figures.
to balance,
m1 x1 = m2 x2
so,
(50.0)(1.90) = (45.0)(x2)
cinchy, right?
thank you so much !!
To find the value of x2 required for the system to be in rotational equilibrium, we need to use the principle of moments.
The principle of moments states that for a system to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.
In this case, we have two persons sitting on opposite sides of the board. The distance of the first person from the fulcrum is given as 1.90 m to the left of the fulcrum. The distance of the second person from the fulcrum is x2 to the right of the fulcrum.
The clockwise moment due to the first person is given by:
Clockwise Moment1 = (m1) * (distance1)
Clockwise Moment1 = (50.0 kg) * (1.90 m)
The counterclockwise moment due to the second person is given by:
Counterclockwise Moment2 = (m2) * (distance2)
Counterclockwise Moment2 = (45.0 kg) * (x2)
To achieve rotational equilibrium, Clockwise Moment1 should be equal to Counterclockwise Moment2. Therefore, we have the equation:
(50.0 kg) * (1.90 m) = (45.0 kg) * (x2)
Now let's solve the equation for x2:
x2 = (50.0 kg * 1.90 m) / (45.0 kg)
Calculating the value:
x2 = 2.111 m
Therefore, the value of x2 required for the system to be in rotational equilibrium is 2.11 m (rounded to 3 significant figures).