# Torque and Tension

1. Oct 21, 2004

### Enoch

Guys - I'm back again :rofl: . I hate to have so many questions, but I really appreaciate all the help you guys give me. For this problem, I am having some problems with the clockwise/counterclockwise Torque rule. Anyways, let me give you the problem and work.

Two weights attached to a uniform beam of mass 26 kg are supported in a horizontal position by a pin and cable as shown in the figure (I have placed the figure in the attachment doc. 1). The acceleration of gravity is 9.8m/s^2. What is the tension in the cable which supports the beam? Answer in units of kN.

To solve this problem, I knew I had to apply the rule that clockwise torques are equal to counterclockwise torques.

Clockwise torques: Weight of the bar, Weight 1, and Weight 2.

Counterclockwise torques: Tension

Here is where my work gets shady . For the tensions, I broke the tension line up into components Ty and Tx. Ty = Tsin (theta). Tx = Tcos (theta). Do I take both of these for the counterclockwise torques in the equation? When I set the problem up, I only used the Ty value. Is this incorrect?

Anyways here is the final problem set up as I had it with only one of the Tension components:

[Weight of the bar X (length/2)] + [Weight1 X distance from weight] + [Weight2 X length] = Tsin (theta) X length

In numerical values: 26(5.6/2) + 32(3.1) + 37(5.6) = Tsin (39) X 5.6

Obviously this answer is wrong, but my question is....is the correct equation this:

[Weight of the bar X (length/2)] + [Weight1 X distance from weight] + [Weight2 X length] = [Tsin (theta) X length] + [Tcos (theta) X length]

In numerical values: 26(5.6/2) + 32(3.1) + 37(5.6) = [Tsin (39) X 5.6] + [Tcos (39) X 5.6]

OR is it something else entirely? Thanks for all the help guys!

#### Attached Files:

• ###### Doc1.doc
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2. Oct 21, 2004

### bem

the last equation should be, for the x-component of T, the distance is 0, not 5.6. Only y-component of T has the distance 5.6 . So u just cancel out x-component cuz it's direction goes through the point O.

3. Oct 22, 2004

### Enoch

Would that not leave me with the exact same initial equation that I started off with? I used that equation and came up with the wrong answer. Either I made a careless mistake (likely), I'm not understanding you, or there is something else wrong with the problem.

4. Oct 24, 2004

### Enoch

bump....Still need help :grumpy: