What Forces are Involved in Torque and Equilibrium Problems?

AI Thread Summary
The discussion focuses on solving torque and equilibrium problems involving forces acting on a system. Key equations are derived for summing forces in both the x and y directions, as well as for calculating torques about a pivot point. Participants emphasize the importance of using weight instead of mass in calculations and ensuring all forces are accounted for, including tension, normal force, and weights of the strut and crate. There is clarification on the correct interpretation of gravitational force (g) in calculations. The conversation highlights the necessity of refining equations to achieve accurate results in the context of torque and equilibrium.
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Homework Statement


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The Attempt at a Solution


The arrows on the diagram are drawn in, as are the labels (except for the angles).
\sum F_{x} = P_{x} - Tcos30 = 0
T = \frac{P_{x}}{cos30}
\sum F_{y} = N_{p} - 8900N - 650g - Tsin30 = 0
N_{p} = 15274N + Tsin30
N_{p} = 15274N + P_{x}tan30

I think I'm missing forces, and probably making up a few. If anyone could check them over before I continue I'd really appreciate it.
 
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you're off to a good start. make sure you have the weight of the box in kg. now you need a third equation, which comes from summing the torques about the pivot and setting that equal to 0.
 
Are all the forces right though? Am I missing any?
Here's the third equation:
The angle between the strut and the tension from the cable is 16 degress: 180 - 30 - (180 - 46) = 16
\sum\tau = Tlsin16 - \frac{1}{2}l650g(cos46) - 8900l(cos46) = 0
\frac{P_{x}}{cos30}lsin16 = 325lg(cos46) + 8900l(cos46)
\frac{P_{x}}{cos30}sin16 = 325g(cos46) + 8900(cos46)
Px = \frac{325g(cos46) + 8900(cos46)}{sin16} = 26381N
T = \frac{26381}{cos30} = 30462N
N_{p} = 15274 + P_{x}tan30 = 15274 + 26381(tan30) = 30505N
P_{tot} = 40330N
\phi = 49 degrees
 
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i just noticed. in your summation of the forces in the y direction, you have the mass of the strut. mass is not a force, so it shouldn't be included in the summation of the forces. you have to find the weight of the strut. (i made a bad mistake by saying that the weight of the crate should be in kg. mass and weight are not the same thing). same deal in your summation of torques. torque is defined as the force times the moment arm, not the mass times moment arm. i think you should have it though. just play around with your three equations till you get it.
 
When I listed 600g, I meant the g as in 9.81 not grams or kilograms. When I computed the answers, I multiplied things by 9.81 wherever g is. I use g so that I get a more accurate answer. Am I using all the right forces though? Tension in cable pulling down and to the left, normal force from pivot on the strut pushing up and to the right, weight of strut going down, and weight of crate going down. Is that all?
 
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