Torque Experiment: Understanding Weight and Torque Relationship

[Meteo]

In this experiment, we had a half meterstick with a hole at one end. This was the pivot point.

Then about 4 cm from the pivot point was a scale which suspended the meterstick horizontally.

Then we had an unknown weight at the other end of the meterstick and we measured the force on the scale with different lever arm lengths of the unknown weight.

Then we graphed the torque exerted on the scale vs the lever arm length of the unknown weight.

The torque on the scale should be the torque of the meter stick + torque of the unknown weight.
the slope should = the weight of the unknown mass

the intercept should be the torque of the meterstick

Why is the slope = to the weight of the unknown mass and why is the intercept = to the torque of the meterstick

All I can do is derive the formula for the weight of the unknown mass in terms of the torque of the scale...
$$\tau_m+\tau_u=\tau_s$$

$$\tau_u=\tau_s-\tau_m$$

$$r_uw_u=\tau_s-\tau_m$$

$$w_u = \frac{\tau_s-\tau_m}{r_u}$$

but that doesn't help me at all...

 

[lippyka]

The slope of the graph is equal to the weight of the unknown mass because the torque of the unknown mass is directly proportional to the lever arm length. As the lever arm length increases, the torque exerted by the unknown mass also increases, and thus the slope of the graph is equal to the weight of the unknown mass. The intercept of the graph is equal to the torque of the meterstick because the torque of the meterstick does not depend on the lever arm length. The torque of the meterstick is constant regardless of the lever arm length, and thus the intercept of the graph is equal to the torque of the meterstick.

 

[quicknote]

I would first like to commend you on conducting a well-designed experiment to understand the relationship between weight and torque. Your experiment set-up and data analysis are both sound and provide valuable insights into this relationship.

Now, to answer your question about the slope and intercept of the graph. Let's start by discussing what torque is. Torque is a measure of the force that causes an object to rotate around an axis or pivot point. In this experiment, the pivot point is the hole at one end of the half meterstick.

Now, when we apply a force to an object at a certain distance from the pivot point, we create a torque. The magnitude of this torque is equal to the force multiplied by the distance from the pivot point, also known as the lever arm. In your experiment, the scale is measuring the force applied by the unknown weight, and the lever arm is the distance from the pivot point to the unknown weight.

So, when we graph the torque exerted on the scale vs the lever arm length of the unknown weight, we are essentially plotting the relationship between the force applied (unknown weight) and the distance from the pivot point (lever arm). This relationship can be expressed as a simple equation: \tau = F \times r, where \tau is torque, F is force, and r is the lever arm.

Now, let's look at the slope and intercept of your graph. The slope represents the rate of change between the torque exerted on the scale and the lever arm length of the unknown weight. In other words, it shows how much torque is being exerted for every unit increase in the lever arm length. Since torque is directly proportional to force and lever arm, the slope of the graph will be equal to the force applied by the unknown weight.

Similarly, the intercept represents the value of torque when the lever arm length is zero, which in this case is the torque of the meterstick. This makes sense because when the lever arm length is zero, the unknown weight is directly above the pivot point, and thus, the torque exerted by the unknown weight is also zero. Therefore, the only torque acting on the scale is the torque of the meterstick.

In conclusion, the slope and intercept of your graph represent the force applied by the unknown weight and the torque of the meterstick, respectively. This is in line with the equation: \tau_m + \tau_u = \tau_s, where \tau_m is the