Lab 121: Rotational Static Equilibrium – Forces on a Strut
PHYS 111A-008
Objectives:
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 (θ_{1}=0°)
Part II. The strut in a position with an angle (θ_{1}≠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 (θ_{1}≠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
θ_{1}=0° | W = 0.7574 kg W_{Exp}=0.757kg | L_{1}=0.37 m |
θ_{2}=50° | W_{1}=0.250 kg | L_{2}=0.465 m |
W_{2}=0.250 kg | L_{3}=0.415 m |
Table II
Weight of strut (Al rod) = 0.11338 kg, L = 0.565 m
θ_{1}=30° | W = 0.510 kg W_{Exp}= .505kg | L_{1}=0.37 m |
θ_{2}=80° | W_{1}=0.250 kg | L_{2}=0.465 m |
W_{2}=0.250 kg | L_{3}=0.415 m |
Table III
Weight of strut (Al rod) = 0.11338 kg, L = 0.565 m
θ_{1}=30° | W = 1.06 kg W_{Exp}=1.106kg | L_{1}=0.37 m |
θ_{2}=27° | W_{1}=0.250 kg | L_{2}=0.465 m |
W_{2}=0.250 kg | L_{3}=0.415 m |
Calculations:
Part 1.
τ_{1}=W_{1} L_{1} sin90°=0.250*0.37*1=0.0925 kg.m τ_{2}=W_{2} L_2 sin90°=0.250*0.465*1=0.11625 kg.m τ_Al=W_Al 1/2 Lsin90°=.11338*.5*0.565*1=0.03203 kg.m τ_w=-WL_3 sinθ_2°=-W*0.415*sin(50°)=-0.31791W τ_net=τ_{1}+τ_{2}+τ_Al+τ_w=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.
τ_{1}=W_{1} L_{1} sin(90°+θ_1 )=0.250*0.37*sin(120°)=0.08011 kg.m τ_{2}=W_{2} L_2 sin(90°+θ_1 )=0.250*0.465*sin(120°)=0.100675 kg.m τ_Al=W_Al 1/2 Lsin(90°+θ_1 )=0.250*0.11338*0.5*0.565*sin(120°)=0.006935 kg.m τ_w=-WL_3 sinθ_2°=-W*0.415*sin(80°)=-0.4087W τ_net=τ_{1}+τ_{2}+τ_Al+τ_w=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.
τ_{1}=W_{1} L_{1} sin(90°-θ_1 )=0.250*0.37*sin(60°)=0.08011 kg.m τ_{2}=W_{2} L_2 sin(90°-θ_1 )=0.250*0.465*sin(60°)=0.100675 kg.m τ_Al=W_Al 1/2 Lsin(90°-θ_1 )=0.250*0.11338*0.5*0.565*sin(60°)=0.006935 kg.m τ_w=-WL_3 sinθ_2°=-W*0.415*sin(27°)=-0.1884W τ_net=τ_{1}+τ_{2}+τ_Al+τ_w=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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