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A meter stick is balanced at its center. If a 3 kg mass is suspended at x=0 m, where would you need to place a mass of 5 kg to have the system in equilibrium? Note: The meter stick goes from 0 m to 1 m; therefore its center is at the 0.5 m mark.

A meter stick is balanced at its center. If a 3 kg mass is suspended at x=0 m, where would you need to place a mass of 5 kg to have the system in equilibrium? Note: The meter stick goes from 0 m to 1 m; therefore its center is at the 0.5 m mark.

Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter8: Rotational Motion
Section: Chapter Questions
Problem 82A
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 A meter stick is balanced at its center. If a 3 kg mass is suspended at x=0 m, where would you need to place a mass of 5 kg to have the system in equilibrium? Note: The meter stick goes from 0 m to 1 m; therefore its center is at the 0.5 m mark.

(It is based on this lab https://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html) 

 

EQUILIBRIUM OF A RIGID BODY Objective: In this experiment a balance is used as a rigid body to illustrate the application of the equations of rotational equilibrium. The torque equation will be verified for a balanced system of two or more masses. The torque equation will also be applied to determine the magnitude of an unknown mass. Introduction: If a rigid body is in equilibrium, then the vector sum of the external forces acting on the body yields a zero resultant and the sum of the torques of the external forces about any arbitrary axis is also equal to zero. Net Force = 0 ΣF0 Eq. 1 Sum of forces on the left = Sum of forces on the right Sum of forces upwards = Sum of forces downwards Net Torque = 0 Στ0 Еq. 2 Sum of clockwise torques = Sum of counter-clockwise torques In this experiment force is defined as the mass, m times gravitational acceleration, g: F = mg Eq. 3 where m is in kg and g in m/s² therefore, force F is in newtons N. Torque is a measure of how effective a given force is at twisting or turning the object it is applied to. Torque is defined as the force F times the moment arm or lever arm r of the force with respect to a selected pivot point x. In other words r is the distance from the pivot to the point where a force is applied. If the force is perpendicular to r, then: T = Fr Eq. 4
Transcribed Image Text:EQUILIBRIUM OF A RIGID BODY Objective: In this experiment a balance is used as a rigid body to illustrate the application of the equations of rotational equilibrium. The torque equation will be verified for a balanced system of two or more masses. The torque equation will also be applied to determine the magnitude of an unknown mass. Introduction: If a rigid body is in equilibrium, then the vector sum of the external forces acting on the body yields a zero resultant and the sum of the torques of the external forces about any arbitrary axis is also equal to zero. Net Force = 0 ΣF0 Eq. 1 Sum of forces on the left = Sum of forces on the right Sum of forces upwards = Sum of forces downwards Net Torque = 0 Στ0 Еq. 2 Sum of clockwise torques = Sum of counter-clockwise torques In this experiment force is defined as the mass, m times gravitational acceleration, g: F = mg Eq. 3 where m is in kg and g in m/s² therefore, force F is in newtons N. Torque is a measure of how effective a given force is at twisting or turning the object it is applied to. Torque is defined as the force F times the moment arm or lever arm r of the force with respect to a selected pivot point x. In other words r is the distance from the pivot to the point where a force is applied. If the force is perpendicular to r, then: T = Fr Eq. 4
The unit of torque is Newton-meter, N-m. The sign for torque is defined as positive (+) when rotating in counter-clockwise direction and negative (-) when rotating in a clockwise direction. For this experiment it is not only important to familiarize yourself with the equations, but also with sketching free-body diagrams (FBD). Using FBDS help solve the problems in question easier and faster. Example # 1: Below is an example where the torque equation is applied to prove the system is in equilibrium: Fulcrum 2 1.75 1.5 1.25 1 0.75 0.5 0,25 Meters 0.25 0.5 0.75 1 1.25 1.5 1.75 2 Meters m,*g M*g m,*g Fig. 1. System in equilibrium and corresponding FBD Since Et = 0 and t = F:r we can develop the equation of equilibrium based on the FBD taking into account the signs of the torques. r is the distance from the fulcrum or pivot point to the point where the force is applied. In the case above we have F1 applied at a moment arm rị that will cause a counterclockwise rotation with respect to the pivot point, therefore, the torque is positive. F2 is acting at a moment arm r2 that will cause a clockwise rotation with respect to the pivot point, thus, making the torque negative. Therefore, Et = T, – T, = 0 T1 = T2 Rewriting in terms of F and r: (F,)(ri) = (F2)(r2) where F=mg therefore, (m1 9)(ri) = (m2g)(r2) Еq. 5 In the system we have: m1= 10 kg, rı= 1.5 m; m2= 15 kg, r2= 1.0 m. Substituting the given values along with g=9.8 m/s² into Eq. 5: (10 kg * 9.8 m/s²)(1.5 m) = (15 kg * 9.8 m/s²)(1 m) 147 N :т %3 147 N :т Therefore, the net torque: Et = 147 N ·m – 147 N ·m = 0 The system is in equilibrium.
Transcribed Image Text:The unit of torque is Newton-meter, N-m. The sign for torque is defined as positive (+) when rotating in counter-clockwise direction and negative (-) when rotating in a clockwise direction. For this experiment it is not only important to familiarize yourself with the equations, but also with sketching free-body diagrams (FBD). Using FBDS help solve the problems in question easier and faster. Example # 1: Below is an example where the torque equation is applied to prove the system is in equilibrium: Fulcrum 2 1.75 1.5 1.25 1 0.75 0.5 0,25 Meters 0.25 0.5 0.75 1 1.25 1.5 1.75 2 Meters m,*g M*g m,*g Fig. 1. System in equilibrium and corresponding FBD Since Et = 0 and t = F:r we can develop the equation of equilibrium based on the FBD taking into account the signs of the torques. r is the distance from the fulcrum or pivot point to the point where the force is applied. In the case above we have F1 applied at a moment arm rị that will cause a counterclockwise rotation with respect to the pivot point, therefore, the torque is positive. F2 is acting at a moment arm r2 that will cause a clockwise rotation with respect to the pivot point, thus, making the torque negative. Therefore, Et = T, – T, = 0 T1 = T2 Rewriting in terms of F and r: (F,)(ri) = (F2)(r2) where F=mg therefore, (m1 9)(ri) = (m2g)(r2) Еq. 5 In the system we have: m1= 10 kg, rı= 1.5 m; m2= 15 kg, r2= 1.0 m. Substituting the given values along with g=9.8 m/s² into Eq. 5: (10 kg * 9.8 m/s²)(1.5 m) = (15 kg * 9.8 m/s²)(1 m) 147 N :т %3 147 N :т Therefore, the net torque: Et = 147 N ·m – 147 N ·m = 0 The system is in equilibrium.
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