A massless bar of length L = 4.0 m is placed on a pivot under its centre. A block of mass m1 4.4 kg is placed at a distance of ri 0.50 m to || the left of the pivot. A second block of mass m2 1.2 kg is placed at a distance of r2 1.5 m to the right of the pivot. (a) What is the net torque due to the masses m1 and m2? (b) Where must a third block of mass m3 = 1.1 kg be placed for the bar to be in static equilibrium? Note: static equilibrium means that the net torque is zero Take g = 9.81 m/s² and show all your work. L r2 mị m2

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Chapter1: Units, Trigonometry. And Vectors
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1. A massless bar of length L = 4.0 m is placed on a pivot under its centre.
A block of mass m1
4.4 kg is placed at a distance of ri
0.50 m to
the left of the pivot. A second block of mass m2
1.2 kg is placed at a
distance of r2 = 1.5 m to the right of the pivot.
(a) What is the net torque due to the masses m1 and m2?
(b) Where must a third block of mass m3
1.1 kg be placed for the bar
to be in static equilibrium? Note: static equilibrium means that
the net torque is zero
Take g = 9.81 m/s? and show all your work.
ri
r2
m1
m2
Transcribed Image Text:1. A massless bar of length L = 4.0 m is placed on a pivot under its centre. A block of mass m1 4.4 kg is placed at a distance of ri 0.50 m to the left of the pivot. A second block of mass m2 1.2 kg is placed at a distance of r2 = 1.5 m to the right of the pivot. (a) What is the net torque due to the masses m1 and m2? (b) Where must a third block of mass m3 1.1 kg be placed for the bar to be in static equilibrium? Note: static equilibrium means that the net torque is zero Take g = 9.81 m/s? and show all your work. ri r2 m1 m2
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