From Michael Herrin, from Chestatee High School in Gainesville, GA:

I’m having trouble working a problem. It’s from the Cutnell and Johnson’s book chapter 4, and I am curious if you could help point me in the right direction.

Mr. Herrin actually pointed me to a problem in the 5th edition, but I found a slightly different version in the newer edition. In the diagram shown, all pulleys are massless and the surfaces are frictionless. Find the tension in the rope and the acceleration of the masses.

The problem solving process that I teach for Newton’s second law problems is:

1. Draw a free body diagram.

2. Break angled forces into components, if necessary.

3. Write (up-down=ma) and (left-right=ma).[1]

This is a two-body problem; therefore, we start with not one but TWO free body diagrams: Look here -->

Note the tricksiness of the diagram for the 3 kg mass. The rope is pulling up TWICE on this mass: once from the left side of the pulley, once from the right side. So we put two tensions on the diagram.

Jacobs’ law of tensions says “One Rope Equals One Tension.” That’s why I didn’t label the tensions T

Now, write Newton’s second law twice, once for each block. The direction of acceleration will be to the right for the 10 kg block, and down for the 3 kg block. (We can see that by imagining releasing the masses from rest, and seeing which way the blocks speed up.)

T = (10 kg)a and (3 kg)g - (T+T) = (3 kg)a

Now, the PHYSICS IS DONE: I have two equations and two unknowns, a and T. Everything else is a known value. All that’s remaining is mathematics.

I think it’s easiest to solve by addition – multiply the first equation by 2, and add the equations together. This cancels out the tension terms; solve to get a = 1.3 m/s

Now, ask your students: what would happen to the acceleration if we were to give the 10 kg mass an initial shove to the left. Would the acceleration be greater than, less than, or equal to 1.3 m/s

GCJ

[1] Well, okay, sometimes we write (right-left=ma), not (left-right=ma). How to know which is which? Determine the direction of acceleration, and start with that direction.

I’m having trouble working a problem. It’s from the Cutnell and Johnson’s book chapter 4, and I am curious if you could help point me in the right direction.

Mr. Herrin actually pointed me to a problem in the 5th edition, but I found a slightly different version in the newer edition. In the diagram shown, all pulleys are massless and the surfaces are frictionless. Find the tension in the rope and the acceleration of the masses.

The problem solving process that I teach for Newton’s second law problems is:

1. Draw a free body diagram.

2. Break angled forces into components, if necessary.

3. Write (up-down=ma) and (left-right=ma).[1]

This is a two-body problem; therefore, we start with not one but TWO free body diagrams: Look here -->

Note the tricksiness of the diagram for the 3 kg mass. The rope is pulling up TWICE on this mass: once from the left side of the pulley, once from the right side. So we put two tensions on the diagram.

Jacobs’ law of tensions says “One Rope Equals One Tension.” That’s why I didn’t label the tensions T

_{1}and T_{2}, but just T. The tension will be the same throughout.Now, write Newton’s second law twice, once for each block. The direction of acceleration will be to the right for the 10 kg block, and down for the 3 kg block. (We can see that by imagining releasing the masses from rest, and seeing which way the blocks speed up.)

T = (10 kg)a and (3 kg)g - (T+T) = (3 kg)a

Now, the PHYSICS IS DONE: I have two equations and two unknowns, a and T. Everything else is a known value. All that’s remaining is mathematics.

I think it’s easiest to solve by addition – multiply the first equation by 2, and add the equations together. This cancels out the tension terms; solve to get a = 1.3 m/s

^{2}? . Plugging back into either equation, I get T = 13 N. That’s reasonable: the tension is less than the weight of either mass, but is on the same order of magnitude.Now, ask your students: what would happen to the acceleration if we were to give the 10 kg mass an initial shove to the left. Would the acceleration be greater than, less than, or equal to 1.3 m/s

^{2}? Post YOUR answer in the comment section.GCJ

[1] Well, okay, sometimes we write (right-left=ma), not (left-right=ma). How to know which is which? Determine the direction of acceleration, and start with that direction.

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