PHYS 111A-008 Lab 121: Rotational Static Equilibrium

Lab 121: Rotational Static Equilibrium – Forces on a Strut

PHYS 111A-008

Objectives:

  1. To determine the tension on the supporting cord and a strut by taking torques about the pivot point of the strut;
  2. To get a better understanding of torque and the condition for rotational equilibrium.

Introduction:

By definition torque is a cross product of distance r ⃗ and applied force( F) ⃗.

(1) τ ⃗=r ⃗ x F ⃗

Magnitude of torque, is:

(2)τ=rFsinθ

When a body is in rotational equilibrium, the sum of all torques which is net torque acting on the body about any point, O, must be zero, that is:

(3) τ_net=∑τ=0

Experimental Procedure:

Set up the assembly. There are two angles in this system, the is the angle between the strut (aluminum bar) and horizontal direction and the is the angle between the strut and the supporting cord.

Part I. The strut in horizontal position (θ1=0°)

  1. Measure L1,L2,L3,and L record the values in data table I. Also record the weight of the aluminum rod of the strut which is written on it.
  2. In order to keep θ1=0°, θ2 has to be a certain value. Measure and record.
  3. Given W1 and W2, calculate the tension W in the supporting cord that could keep the strut in a horizontal position. Show your calculations.
  4. Set up the apparatus and measure the tension in the supporting cord necessary to maintain equilibrium.
  5. Compare your experimental results with the calculations.

Part II. The strut in a position with an angle (θ1≠0°) tilted up

Repeat the procedure as in Part I. Record the values in data table II.

Part II. The strut in a position with an angle (θ1≠0°) tilted down

Repeat the procedure as in Part I. Record the values in data table III.

Data Tables

Table I

Weight of strut (Al rod) = 0.11338 kg, L = 0.565 m

θ1=0° W = 0.7574 kg WExp=0.757kg L1=0.37 m
θ2=50° W1=0.250 kg L2=0.465 m
W2=0.250 kg L3=0.415 m

Table II

Weight of strut (Al rod) = 0.11338 kg, L = 0.565 m

θ1=30° W = 0.510 kg WExp= .505kg L1=0.37 m
θ2=80° W1=0.250 kg L2=0.465 m
W2=0.250 kg L3=0.415 m

Table III

Weight of strut (Al rod) = 0.11338 kg, L = 0.565 m

θ1=30° W = 1.06 kg WExp=1.106kg L1=0.37 m
θ2=27° W1=0.250 kg L2=0.465 m
W2=0.250 kg L3=0.415 m

Calculations:

Part 1.

τ1=W1 L1 sin90°=0.250*0.37*1=0.0925 kg.m
τ2=W2 L2 sin90°=0.250*0.465*1=0.11625 kg.m
τAl=WAl  1/2 Lsin90°=.11338*.5*0.565*1=0.03203 kg.m
τw=-WL3 sinθ2°=-W*0.415*sin⁡(50°)=-0.31791W
τ_net=τ12Alw=0
0.24078-0.31791W=0
W_theoratical=0.7574 kg
Percent error=(|Theoratical-Experimental|)/Theoratical*100%
=|0.7574-0.757|/0.7574*100%
=0.05%

Part 2.

τ1=W1 L1  sin⁡(90°+θ_1 )=0.250*0.37*sin⁡(120°)=0.08011 kg.m
τ2=W2 L2  sin⁡(90°+θ_1 )=0.250*0.465*sin⁡(120°)=0.100675 kg.m
τAl=WAl  1/2 Lsin(90°+θ_1 )=0.250*0.11338*0.5*0.565*sin⁡(120°)=0.006935 kg.m
τw=-WL3 sinθ2°=-W*0.415*sin⁡(80°)=-0.4087W
τ_net=τ12Alw=0
0.2085-0.4087W=0
W_theoratical=0.510 kg
Percent error=(|Theoratical-Experimental|)/Theoratical*100%
=|0.510-0.505|/0.510*100%
=1%

Part 3.

τ1=W1 L1  sin⁡(90°-θ_1 )=0.250*0.37*sin⁡(60°)=0.08011 kg.m
τ2=W2 L2  sin⁡(90°-θ_1 )=0.250*0.465*sin⁡(60°)=0.100675 kg.m
τAl=WAl  1/2 Lsin(90°-θ_1 )=0.250*0.11338*0.5*0.565*sin⁡(60°)=0.006935 kg.m
τw=-WL3 sinθ2°=-W*0.415*sin⁡(27°)=-0.1884W
τ_net=τ12Alw=0
0.2085-0.1884W=0
W_theoratical=1.1068 kg
Percent error=(|Theoratical-Experimental|)/Theoratical*100%
=|1.1068-1.106|/1.1068*100%
=0.07%

Conclusion:

This lab helped us learn how to find torque by using rotational static equilibrium; it also helped us prove that the net torque in the system should equal zero. This lab turned out to be quite successful considering the low amount of percentage error we got from our calculations. Even though in this lab there were a lot of factors that could have caused errors. But we managed to get accurate results that matched theoretical values we calculated.

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