Enoch
Oct21-04, 08:26 PM
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 :confused: . 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!
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 :confused: . 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!