Question

7. Karen is trying to make a mobile. She wants to balance two masses hanging from a stick. The stick is also hanging by a string tied around the center of the stick from the ceiling. If one mass has a weight of 5.3 N and it is 0.37 m from the middle of the stick. How far should the second mass be placed from the middle on the other side if it has a weight of 9.5 N?

Answer 1

I think you'll need to use torque

\(τ=rFsin\theta\)
Let mass one be A.
\(τ_a = r_a F_a sin(\theta)\)
We must assume the degree is perpendicular, so thus theta is 90. sin90 is just 1.
Radius of a is 0.37m
Force of a is 5.3 N
so \(τ_a=0.37m*5.3N=1.961Nm\)

Now \(τ_b\) of mass b must be the same as the torque of a, or 1.961 Nm
They are giving weight which is just a force
You must find the radius
\(τ_b=r_b F_b\)
\(1.961Nm = r_b 9.5N\)
\(r_b=0.205m\)

Answer is 0.205 meters.

Answer 2

Thank you so much!!

Answer 3

NP