Two-dimensional Equilibrium

Name:

Lab Partners:

When entering numeric data, use exponentials: ie., 1.6 * 10^-19 = 1.6E-19.

Procedure

1. Assemble the force table and level it. Measure the masses of the weight hangers:

   mh, 1 = g				mh, 3 = g
   mh, 2 = g

2. Set two pulleys at 35 and 140 degrees and place 150g on each of the corresponding weight hangers. Using a third pulley, hanger and weights, determine the position ( q_i ) and additional weight ( m_i ) necessary to balance the ring in a position centered over the central pin. Pull the ring up a short distance and release it several times in order to minimize the effects of friction in the pulleys.

   The masses below are those added to the weight hangers!

   m1 = _________150________ g				q1 = _________35_________ degrees
   m2 = _________150________ g				q2 = ________140_________ degrees
   m3 = g				q3 = degrees

3. Repeat step 2 for 200g at 25 degrees and 150g at 95 degrees:

   m1 = _________200________ g				q1 = _________25_________ degrees
   m2 = _________150________ g				q2 = _________95_________ degrees
   m3 = g				q3 = degrees

4. Repeat step 2 for 200g at 0 degrees and 150g at 90 degrees:

   m1 = _________200________ g				q1 = __________0_________ degrees
   m2 = _________150________ g				q2 = _________90_________ degrees
   m3 = g				q3 = degrees

5. Place 200g on a weight hanger at 220 degrees. Place the other two weight hangers at 0 and 90 degrees and determine how much weight is required on each to balance the center ring as in step 2.

   m1 = _________200________ g				q1 = ________220_________ degrees
   m2 = g				q2 = __________0_________ degrees
   m3 = g				q3 = _________90_________ degrees

Analysis

1. For each of the configurations in steps 2 through 5 above, draw a vector diagram on graph paper. The direction will be the same as during the experiment and the length will correspond to the weight (including that of the weight pan).
2. Compute the x and y components of all force vectors ( f = m g ) in all four configurations:

   F_{x, i} = ( m_i + m_{h, i} ) ( 981 cm / s² ) cos ( q_i )

   F_{y, i} = ( m_i + m_{h, i} ) ( 981 cm / s² ) sin ( q_i )

   from step 2:

   Fx, 1 = dynes				Fy, 1 = dynes
   Fx, 2 = dynes				Fy, 2 = dynes
   Fx, 3 = dynes				Fy, 3 = dynes

   from step 3:

   Fx, 1 = dynes				Fy, 1 = dynes
   Fx, 2 = dynes				Fy, 2 = dynes
   Fx, 3 = dynes				Fy, 3 = dynes

   from step 4:

   Fx, 1 = dynes				Fy, 1 = dynes
   Fx, 2 = dynes				Fy, 2 = dynes
   Fx, 3 = dynes				Fy, 3 = dynes

   from step 5:

   Fx, 1 = dynes				Fy, 1 = dynes
   Fx, 2 = dynes				Fy, 2 = dynes
   Fx, 3 = dynes				Fy, 3 = dynes

3. For steps 2 through 4, compute the absolute and relative errors between the magnitudes of the x components of 1) the sum of the given vectors and 2) the resultant vector which you constructed experimentally:

   F_{x, abs} = | | F_{x, 1} + F_{x, 2} | - | F_{x, 3} | |.

   When computing the relative error, use the given vector in the denominator:

   F_{x, rel} = 100 % x F_{x, abs} / | F_{x, 1} + F_{x, 2} |.

   This implies that the given vectors is the standard to which the measured vector is being compared. Repeat for the y components.

   from step 2:

   Fx, abs = dynes				Fy, abs = dynes
   Fx, rel = %				Fy, rel = %

   from step 3:

   Fx, abs = dynes				Fy, abs = dynes
   Fx, rel = %				Fy, rel = %

   from step 4:

   Fx, abs = dynes				Fy, abs = dynes
   Fx, rel = %				Fy, rel = %

4. For step 5, compute the absolute and relative errors between the x components of 1) the given vector and 2) the sums of the experimentally determined vectors. Repeat for the y components.

   F_{x, abs} = | | F_{x, 1} | - | F_{x, 2} + F_{x, 3} | |.

   F_{x, rel} = 100 % x F_{x, abs} / | F_{x, 2} + F_{x, 3} |.

   Fx, abs = dynes				Fy, abs = dynes
   Fx, rel = %				Fy, rel = %

Does your data support the statement that in equilibrium, the sum of the x components of the forces must equal zero (and the same, independently, for the y components) ?