Suppose an experimental arrangement like the one in the table has mass m1 = 0.200 kg at the 0.100-m mark and a mass m2 = 0.100 kg at the 0.750-m mark. Can the system be put in equilibrium by a 0.050-kg mass? If it can be done, state where it would be placed. If it cannot be done, state why not.

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Chapter1: Units, Trigonometry. And Vectors
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Suppose an experimental arrangement like the one in the table has mass m1 = 0.200 kg at the 0.100-m
mark and a mass m2 = 0.100 kg at the 0.750-m mark. Can the system be put in equilibrium by a
0.050-kg mass? If it can be done, state where it would be placed. If it cannot be done, state why not.

Mass (kg)
m₁=0.100
m₂=0.200
m3 0.050
=
Position (m)
x₁=0.100
x₂=0.750
x3 = 0.405
Lever arm (m)
d₁ = 0.399
d₂= 0.251
d3= 0.94
Torque (N-m)
0.391
Tccw
-
Tcw = 0.461
% Difference
16.43%
Transcribed Image Text:Mass (kg) m₁=0.100 m₂=0.200 m3 0.050 = Position (m) x₁=0.100 x₂=0.750 x3 = 0.405 Lever arm (m) d₁ = 0.399 d₂= 0.251 d3= 0.94 Torque (N-m) 0.391 Tccw - Tcw = 0.461 % Difference 16.43%
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