Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

EEK 1010 Chapter 8 Activity: Torques When forces are applied to a rigid body, if the body is to be in equilibrium, it must be

image text in transcribedimage text in transcribedimage text in transcribed
image text in transcribedimage text in transcribedimage text in transcribed
EEK 1010 Chapter 8 Activity: Torques When forces are applied to a rigid body, if the body is to be in equilibrium, it must be so both with respect to translation and rotation. Part 1: Torque due to Two Known Forces We use a meter stick of mass m.J = 0.090 kg and center of gravity at 3,, = 0.50 m. 1. With the meter stick supported at the equilibrium position 3,3, we placed a mass m] = 0.1000 kg at the 0.1000 m mark. We then determined experimentally, to the nearest 0.5 mm, the position X2 at which a mass m; = 0.2000 kg must be placed in order to balance the meter stick. This is recorded as x; in Data Table 1. 2. Calculate the lever arm di for each force. For this case di = | 2% Xi |, where xi is the position of the ithmass. Note that since the support is at the position of the center of gravity ggg, the meter stick mass has zero leverarm and thus will not contribute to the torque. Record the values of d] and d2 in Calculation Table 1. 3. Calculate and record in Calculations Table 1 the value for the sum of the counterclockwise torques 21 tm = m1 gd1. Calculate and record in Calculations Table 1 the value for the sum of the clockwise torques Z "cw = ngdz. 4. According to the theory, the magnitude of Z remand Z 1% should be equal since the system is in equilibrium. Calculate the percentage difference between 2 rm and Z "cw and record the result in Calculations Table l. 4 d] bd d2 b 0 X1 3% X2 100 cm Part 2: Torque due to Three Known Forces 1. Once again the meter stick is supported at its center of gravity 33, We placed masses m1 = 0.1000 kg at the 0.1000 m mark and m; = 0.2000 kg at the 0.7500 m mark. We then determine experimentally, to the nearest 0.5 mm, the position X3 at which a mass m; = 0.0500 kg must be placed in order to balance the system. This is record the value of X3 in Data Table 2. For this experimental arrangement, the meter stick mass m.J again makes no contribution to the torque. Calculate the lever arm for each of the other masses and record the values in Calculations Table 2 (di = | gggl xi I). Calculate the values of Z wand of 2 1% and record these in the Calculations Table 2. Notice that you must decide for mass m; whether it contributes to the counterclockwise or clockwise torque. Calculate the percentage difference between 2 Imand 2 twand record it in Calculations Table 2. 0 X1 X3 gggl X2 100 cm Summary Questinnsl 1. F or both parts of the experiment, when the forces are balanced, discuss the agreement between 2 \"cm and E Tow. Are the experimental results sufciently close to consider that the data veries the theory? Explain your answer. 2. In part 1 of the laboratory, what is the value of the force Fs with which the support pushes upward on the meter stick?I Show your work. METER STICK DATA TABLE m: 0.090 kg 3g = 0.50 In + DATA AND CALCULATIONS TABLE 1 m1 =0.1162 x1 =0.1000 nu=0.2162 x2=0.?150 DATA AND CALCULATIONS TABLE 2 m1= 0.1162 x1 = 0.1000 n12 = 0.2162 x2 = 0.7500 m3 = 0.0662 x3 = 0.3840 Notice: the mass of each hanger is 0.0162 kg

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Advanced Strength and Applied Elasticity

Authors: Ansel C. Ugural, Saul K. Fenster

4th edition

130473929, 130473928, 978-0130473929

More Books

Students also viewed these Physics questions

Question

Always show respect for the other person or persons.

Answered: 1 week ago