Question

Complete the following steps:
• TWO APPLIED FORCES

– Place a pulley at the 20◦ mark on the force table and place a total of 0.1 kg(including the mass holder) on the end of the string. Calculate the magnitude of the force produced by the mass using g = 9.8 m/s2.Assume three significant figures for this and all other calculations of
force. Record the value of this force as F1.

– Place a second pulley at the 90◦ mark on the force table and place a
total of 0.2 kg on the end of the string. Calculate the force produced
and record as F2.

– Determine by trial and error the magnitude of the mass needed and angle at which it must be located for the ring to be centered on the force table. Jiggle the ring slightly to be sure that this equilibrium condition is met. Attach all strings to the ring so that they are
directed along a line passing through the center of the ring. All the forces will then act through the point at the center of the table. Record this value of mass as mFE1.

– Calculate the force produced on the experimentally determined mass.
Record the magnitude and direction of this equilibrium force FE1.

– The resultant force FR1 is equal in magnitude to FE1, and its direc-
tion is 180◦ from FE1. Record the magnitude of the force FR1, the mass equivalent of this force, and the direction of the force.

• THREE APPLIED FORCES

– Place a pulley at 30◦ with 0.15 kg onit, one at 100◦ with 0.2 kg and one at 145◦ with 0.1 kg on it.

– Calculate the force produced by those masses and record them as F3,
F4, and F5.

– Determine the equilibrium force and the resultant force by following a procedure like that followed earlier above. Record the magnitudes of the forces, the associated values of mass, andthe directions.

Calculations

Complete the following calculations:
• TWO APPLIED FORCES

– Use trigonometry to calculate the components of F1 and F2 and record them. Add the components algebraically and determine the magnitude of the resultant by the Pythagorean theorem. Determine the angle of the resultant from the arc tan of the components. Record those results.

– Calculate the percentage error of the magnitude of the experimental
of FR compared to the analytical solution of FR.

• THREE APPLIED FORCES

– Use trigonometry to calculate the components of all three forces, the
components of the resultant, and the magnitude and direction of the resultant, and record them.

– Make the same error calculation for this problem as described earlier
above. Record the value.

• To determine the force acting on each mass it was assumed that g =9.8 m/s2. The value of g at the place where the experiment is performed may be slightly different from that value. State what effect (if any) it would have on the percentage error calculated for the comparisons.

• Two forces are applied to the ring of a force table, one at 80◦, and the other at 20◦. Regardless of the magnitudes of the forces, in which quadrant will the equilibrium be found? and in which quadrant will the resultant be found?

• Pulleys introduce a possible source of error because of their possible fric-
tion. Given that they are a source of error, why are the pulleys used at all? WHat is the function of the pulleys?

I have no idea what to do here... all I got was the F1 = 0.981 N and F2 = 1.962... F1 = Cos(30) / F2 = Sin(80) and F3=Sin(35) ...

I believe the final result is in quadrant III? 248.2? Can anyone help me out with what to do here?

Answer 1

2 applied forces

F1 = 0.1g = 0.98 N at 20^o

x-comp F_1x = 0.98 Cos(20) = 0.921 N

y-comp F_1y = 0.98Sin(20) = 0.335 N

F2 = 0.2g = 1.96 N at 90 deg

x-comp = 1.96 Cos(90) =0

y-comp = 1.96 Sin(90) = 1.96

adding x and y components

resultanat x-comp = 0.921 ; y-comp = 1.96 +0.335 = 2.295

magnitude = sqrt(0.921² + 2.295²) = 2.473 N

direction = arcTan(y/x) = arcTan(2.295/0.921) = 68.13 deg with x-axis

3- force

F3 = 0.15g = 1.47 N at 30 deg

x-comp = 1.47Cos(30) = 1.27N ; y-comp = 1.47Sin(30) = 0.735 N

F4 = 0.2g = 1.96g at 100 deg

x-comp= 1.96Cos(100) = -0.34 N ; y-comp = 1.96Sin(100) = 1.93 N

F5 = 0.1g = 0.98N at 145 deg

x-comp = 0.98Cos(145) = -0.80 N ; y-comp = 0.98Sin(145) = 0.56 N

add up all the x-comp and the y-comps

resultant x-comp = -0.80 - 0.34 +1.27 = 0.13 N

y-comp = 0.56 + 1.93 + 0.735 = 3.225

magnitude = sqrt(0.13² + 3.225² ) = 3.23 N

direction arcTan(y/x) = arcTan(3.225/0.13) = 87.69 deg with x-axis

From the experiment you need to findout the mass to keep the ring in equilibrium and calculate the magnitude of the equilibrium force and its direction , it is equal to the resultant in magnitude and 180 deg in direction to the resultant force.