2. Two masses hang below a mass less meter stick. Mass 1 is located at the 10 cm

Question

2. Two masses hang below a mass less meter stick. Mass 1 is located at the 10 cm mark with a weight of 15 kg, while mass 2 is located at the 60 cm mark with a weight of 30 kg. At what point in between the two masses must the string be attached in order to balance the system?

3. Add the following 3 displacement vectors using the analytical method. report the x-component and y-components of the resultant, the magnitude of the resultant, and the direction of the resultant.
vector A = 1.5 m/s at 20 degrees
vector B = 3 m/s at 60 degrees
vector C = 5 m/s at 210 degrees

Explanation / Answer

Let us suppose the the string be attached at x cm mark.

in order to balance the system, thier torques should be equal from this point(where string is attached)

So balancing the torque.

15g(x-10)=30g(60-x)

x-10=120-2x

3x=130

x=43.33cm

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