Practice Physics Question from Kaplan

[karinab]

Here's the question. How would you approach the problem?

A uniform rod 2 m in length with mass of 4kg is held at an angle A (between 0 and 90 degrees) to a frictionless tabletop by a string pulling up on 0.5m from the upper end of the rod. What is the tension in the string?

There is a diagram that follows but the question describes it accurately. The rod is at an angle A with the tabletop.

The explanation did not make sense to me; help is greatly appreciated!

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[ilovemcat]

Here's the question. How would you approach the problem?

A uniform rod 2 m in length with mass of 4kg is held at an angle A (between 0 and 90 degrees) to a frictionless tabletop by a string pulling up on 0.5m from the upper end of the rod. What is the tension in the string?

There is a diagram that follows but the question describes it accurately. The rod is at an angle A with the tabletop.

The explanation did not make sense to me; help is greatly appreciated!

Draw a diagram first.
Draw the forces acting on your diagram.

At the center (1m) is the downward weight of the metal rod. (Notice they mention the word "uniform," which just means the weight acts at the center of the rod.)
Then they mention a string is attached, "0.5m from the upper end of the rod" which is the same thing as 1.5m the bottom end (where the rod is touching the table). Draw the tension force acting upwards (because it's pulling the rod up).

Now because we don't know the Normal Force of the table, you just approach this as a torque problem.

Take the point where the rod is touching the table to be the pivot point. This cancels out the need for unknown normal force (since no torque can act there).
Because the rod is held in equilibrium, all Upward Forces balance all Downward Forces.

The only two forces labeled on our diagram now is Tension (upwards) and the downward weight of the rod (acting at the center) .

We can plug in these values into the Torque Equation and solve for the unknown (in this case, the Tension of the rod)

Therefore, CW Torque = CCW Torque

Weight x Lever Arm = Tension x Lever Arm
40N x 1m cos(theta) = Tension x 1.5m cos(theta) ==> (Remember, we have to find the Lever Arm)

Tension = 40N x 1m cos(A) / 1.5m cos(A)
Tension = 40 N x 2/3
Tension = 26.67 N

You didn't give an answer, so I hope this is right.

 

[TheRealAngeleno]

Draw a diagram first.
Draw the forces acting on your diagram.

At the center (1m) is the downward weight of the metal rod. (Notice they mention the word "uniform," which just means the weight acts at the center of the rod.)
Then they mention a string is attached, "0.5m from the upper end of the rod" which is the same thing as 1.5m the bottom end (where the rod is touching the table). Draw the tension force acting upwards (because it's pulling the rod up).

Now because we don't know the Normal Force of the table, you just approach this as a torque problem.

Take the point where the rod is touching the table to be the pivot point. This cancels out the need for unknown normal force (since no torque can act there).
Because the rod is held in equilibrium, all Upward Forces balance all Downward Forces.

The only two forces labeled on our diagram now is Tension (upwards) and the downward weight of the rod (acting at the center) .

We can plug in these values into the Torque Equation and solve for the unknown (in this case, the Tension of the rod)

Therefore, CW Torque = CCW Torque

Weight x Lever Arm = Tension x Lever Arm
40N x 1m cos(theta) = Tension x 1.5m cos(theta) ==> (Remember, we have to find the Lever Arm)

Tension = 40N x 1m cos(A) / 1.5m cos(A)
Tension = 40 N x 2/3
Tension = 26.67 N

You didn't give an answer, so I hope this is right.

Hey ilovemcat could you explain why you multiplied the torque by cosA? I'm having a bit of trouble visualizing this. Thanks a lot.

 

[tttgo]

Was that answer correct? My physics isn't very good, but I'll try to help by following my textbook. My book defines torque as the product of the distance r from the rotation axis and the component of the force perpendicular to that axis. So if it's the cross product, shouldn't sin be used since it's the side opposite to the rod. sin(theta)=opposite/hypotenuse. maybe you used a different angle or something. The force balance seems right.

 

[Rabolisk]

The angle between the force and the rod is complementary to the angle between the rod and the table. Drawing the diagram is the best way to visualize this. Since sin(&#952 = cos(90-&#952, ilovemcat used cos(A), which is equal to sin of the angle between the force and the rod.

In the end though, this doesn't matter because both forces are acting on the rod at the same angle, just different direction. So sin or cos function, whichever one you use, is canceled out. What really mattered was that weight was 1 m from the pivot, and that the tension was 1.5 m from the pivot.