For two existing torques, what third force at a given distance from the pivot will balance them?Imagine a meter stick set up as in the figure. It hangs from a central bracket, and two hanging masses can hang from it from each of their brackets. At a third location, a force probe can either pull up or pull down on the stick, depending on what is needed to balance the stick.The mass of the meter stick is 120 g.sketch the situation (drawing r1, r2, r3, F1, F2, and F3) and determine the magnitude (value) and direction (+ or -) of each torque. Don't include the mass of a bracket that would hold the hanging mass in place; assume the mass listed is the entire mass hanging at that point.For each trial, use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x3 Trial1m₂ (g)50x₁ (cm)20m₂ (g)100x₂ (cm)80X3 (cm)60fulcrum location (cm)50Minimummeasured F3 (N)1.32Maximummeasured F3 (N)1.55 m1rIr2VM₂

Name: For two existing torques, what third force at a given distance from th
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Description: For two existing torques, what third force at a given distance from the pivot will balance them?Imagine a meter stick set up as in the figure. It hangs from a central bracket, and two hanging masses can hang from it from each of their brackets. At a

Given: m1 = 50g = 0.05 kg m2 = 100g = 0.1kg x1 = 20 cm = 0.2m x2 = 80 cm = 0.8m therefore r1 = 30cm = 0.3m (x1 is the distance from 0 mark so from fulcrum the location of mass attached is r1) and r2 = 30 cm = 0.3m (x2 is distance from 0 mark so from fulcrum the location of mass attached is r2) fulcrum location 50 cm mass of meter stick is 120gm. as center of mass of the meter stick is at point of pivot/fulcrum it does not produce torque. Torque produced by mass m1; τ1=r1×F=r1×m1g=0.3×0.05×10=0.15 Nm anticlockwise (+ve) Torque produced by mass m2; τ2=r2×F = r2×m2g=0.3×0.1×10=0.3 Nm clockwise (-ve)For equilibrium condition; τ1=τ2 i.e clockwise moments=anticlockwise moments In our case τ2>τ1 i.e clockwise moment is greater than anticlockwise moment. So the meter stick will tend to rotate clockwise. Given x3 = 60cm = 0.6m So r3 = 10 cm = 0.1m So r3*F3 should give 0.15 Nm so that the stick balances. (as need to be applied in anticlockwise ) 0.1*1.32 = 0.07575 0.1*1.55 = 0.155 Nm. So force of 1.55 N will balance the meter stick in such a way that it produces anticlockwise motion. So Force of 1.55N should be applied at x3 = 60cm (or r3=10cm from fulcrum) in vertically upward direction.

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For two existing torques, what third force at a given distance from the pivot will balance them?

Imagine a meter stick set up as in the figure. It hangs from a central bracket, and two hanging masses can hang from it from each of their brackets. At a third location, a force probe can either pull up or pull down on the stick, depending on what is needed to balance the stick.

The mass of the meter stick is 120 g.

sketch the situation (drawing r₁, r₂, r₃, F₁, F₂, and F₃) and determine the magnitude (value) and direction (+ or -) of each torque. Don't include the mass of a bracket that would hold the hanging mass in place; assume the mass listed is the entire mass hanging at that point.

For each trial, use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x₃

Trial
1
m₂ (g)
50
x₁ (cm)
20
m₂ (g)
100
x₂ (cm)
80
X3 (cm)
60
fulcrum location (cm)
50
Minimum
measured F3 (N)
1.32
Maximum
measured F3 (N)
1.55

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