Torque/Static Equilibrium Problems

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SUMMARY

This discussion focuses on solving torque and static equilibrium problems involving forces applied at angles and distances from a pivot point. The first problem involves calculating the magnitude of a force required to maintain equilibrium on a 30cm stick with a 3.3N force applied at an angle of 0.7 radians. The solution yields a required force of 6.844N applied perpendicular to the stick at a distance of 5.0cm from the pivot. The second problem calculates the torque generated by a 2.4N force applied at an angle of 0.5 radians, resulting in a torque of 0.09 Nm.

PREREQUISITES
  • Understanding of torque calculations using the formula Torque = Force * Perpendicular Distance
  • Knowledge of trigonometric functions, specifically sine, for calculating perpendicular distances
  • Familiarity with static equilibrium concepts and conditions for rotational equilibrium
  • Basic physics principles related to forces and moments
NEXT STEPS
  • Study the application of torque in various mechanical systems
  • Learn about static equilibrium conditions in two-dimensional systems
  • Explore advanced torque calculations involving multiple forces and angles
  • Investigate the role of pivot points in rotational dynamics
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This discussion is beneficial for physics students, engineering students, and anyone involved in mechanics or statics who seeks to deepen their understanding of torque and static equilibrium problems.

blackice552
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1. Homework Statement :
I have 2 problems:
1. We have a stick of length 30cm. The stick has a pivot point in its left end. We have a force 3.3N acting at angle 0.7rads and a distance 16.1cm from the pivot. What is the MAGNITUDE of the force we must apply perpendicular to the stick at a distance 5.0cm from the pivot so that the stick does NOT rotate?

2.I am applying a force 2.4N on a wrench at an angle 0.5rads from the horizontal. If the point of application of the force is 7.8cm from the pivot, what is the MAGNITUDE of the torque about the pivot?



2. Homework Equations :
Torque = Force * Perpendicular Distance
Perpendicualr Distance = distance * sin(theta)
Note: I'm not sure if it matters in problem one that the distances aren't the end of the object.

3. The Attempt at a Solution :
1.
Torque1 +Torque2 = 0
Torque1 = .161 m * (3.3 N) * (sin (.7)) = 3.422
Torque2 = -3.422 = F(.5)(sin (pi/2)) = .5F
-6.844 = F
Mag. F = 6.844 N

2.Perpendicular Distance = .078 m * sin(.5) = .037
Torque = 2.4 N * .037 m= .09 Nm
 
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blackice552 said:
Torque2 = -3.422 = F(.5)(sin (pi/2)) = .5F

I'm not sure why you used the sine here, since the distance is perpendicular? Or am I missing something?
 

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