Torque of suspended meter stick

[Punkyc7]

Homework Statement

A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A2.0-N weight is suspended from the 10-cm position on the stick, another 2.0-N weight is suspendedfrom the 50 cm position, and a 3.0-N weight is suspended from the 60 cm position. What is the tension in the string attached at the 100-cm end of the stick?
a. 1.9 N
b. 3.0 N
c. 3.5 N
d. 4.0 N
e. 4.4 N

Homework Equations

sum of torques and sum of forces

The Attempt at a Solution

2*.1+2*.5+3*.6=F(1)

i get the tension to be 3 but the correct one should be 3.5, i don't see where i lose the .5 Newtons

 

[flatmaster]

Does the meter stick have mass? Where does the torque for the stick's mass act?

 

[kerimek]

I would first check the units used in the calculation. It seems that the weights are given in Newtons (N) and the positions are given in centimeters (cm). In order to properly calculate torque, the units must be consistent. Therefore, the positions should be converted to meters (m) before plugging them into the equation.

Using this conversion, the equation would become:

2*0.1m + 2*0.5m + 3*0.6m = F(1)

This would result in a tension of 3.5 N, which matches the correct answer.

Additionally, I would also check the direction of the torque. Since the meter stick is suspended horizontally, the torque due to the weights on either side should be in opposite directions. This means that the torque from the 2.0 N weight at the 50 cm position should be negative, while the torque from the 3.0 N weight at the 60 cm position should be positive.

Taking this into account, the equation would be:

2*0.1m - 2*0.5m + 3*0.6m = F(1)

This would also result in a tension of 3.5 N.

In conclusion, as a scientist, I would carefully check the units and directions involved in the calculation to ensure an accurate and consistent solution.