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A meterstick of uniform density hanging from a string at 0.7 m, had two masses of 0.50 kg and 1.00 kg hung at 0.1 m and 1.0 m respectively to obtain static equilibrium. A calculation for placement of the masses to obtain static equilibrium was done prior to the setup and it came out to: (0.5 kg x 9.8 N/kg x 0.6 m) – (1.00 kg x 9.8 N/kg x 0.3 m) = τnet = 0 N*m However, the setup failed and static equilibrium was not achieved. What do you think went wrong with the setup and/or equation?

User Adnan Ali
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1 Answer

6 votes

Answer:

The weight of the meter stick is missing in the equation and it would have provided counterclockwise torque.

Step-by-step explanation:

Given that,

Length of string = 0.7 m

One mass = 0.50 kg

Another mass = 1.00 kg

Distance = 0.1 m

The given equation is


0.5*9.8*0.6-1.00*9.8*0.3=\tau_(net)=0...(I)

The length of the meter stick is


l = 0.6+0.3-0.7=0.2\ m

According to the question,

The equation will be


0.5*9.8*0.6+w*0.2-1*9.8*0.3=0...(II)

On comparing equation (I) and (II)

So, the missing term is the torque of self weight of stick.

Hence, The weight of the meter stick is missing in the equation and it would have provided counterclockwise torque.

A meterstick of uniform density hanging from a string at 0.7 m, had two masses of-example-1
User Bgh
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