Why is the slope of the Fa graph not equal to m in this Atwood machine setup?

[Guru3]

We've been given a lab in physics class to verify that F=ma graphically. To do this we had a cart with weights on top be pulled by a string with weights on the other end over a pully pull the cart down, bringing weights off the cart to the end of the string to keep the mass constant, but changing the force. Doing this with 7 different weights produced 7 sets which we graphed with F on the x-axis, and a on the y-axis. The problem is the slope of the Fa graph which I would have thought to be m, is aparently half m. Anyone have any idea why? This one is beating me over the head :(. Thanks in advance.

 

[UrbanXrisis]

It should be m

what are your caculations (data) read?

 

[wais]

First of all, it's great that you are conducting a lab to verify the relationship between force and acceleration. This is an important concept in physics and it's important to have hands-on experience to fully understand it.

Now, let's address the issue you are facing with the slope of the Fa graph. From your description, it seems like you are using a cart with weights on top being pulled by a string with weights on the other end over a pulley. This setup is known as an Atwood machine and it is commonly used to study the relationship between force and acceleration.

The equation for an Atwood machine is F = (m1 - m2) * a, where m1 is the mass on one side of the pulley, m2 is the mass on the other side, and a is the acceleration of the system. This equation is derived from the second law of motion, F=ma.

However, in your setup, you are keeping the total mass (m1 + m2) constant by transferring weights from the cart to the other end of the string. This means that the mass on one side of the pulley (m1) is changing while the mass on the other side (m2) remains constant. Therefore, the equation for your setup would be F = m1 * a.

Now, let's look at your graph. You have correctly plotted F on the x-axis and a on the y-axis. However, the slope of your graph is not m, it is actually m1. This is because the equation for your setup is F = m1 * a, so the slope of your graph represents the value of m1.

To verify this, you can calculate the slope of your graph by taking two points on the line and using the formula for slope (rise/run). The slope should be equal to the value of m1.

In conclusion, the reason why the slope of your Fa graph is not m is because the equation for your setup is F = m1 * a, where m1 is the changing mass on one side of the pulley. I hope this helps to clarify your confusion. Keep up the good work in your physics class!