### PHY11L - E204 TORQUE: SECOND CONDITION OF EQUILIBRIUM

 EXPERIMENT 204:  TORQUE: SECOND CONDITION OF EQUILIBRIUM

DATA and OBSERVATIONS

 TABLE 1. Determining the Weight of Pans Actual value of pan 1, =        24.8      grams            Actual value of pan 2, =       24.8      grams Trial     (computed) (computed) 1 =10 g 15    cm 21   cm 24.5    cm 15    cm 25   g 25   g = 5 g 2 =15 g 12.5   cm 20    cm 25    cm 25    cm 25   g 25   g =25 g 3 =30 g 10   cm 22    cm 25    cm 22.5   cm 25   g 25   g =20 g Average weight of pan 1, =          25        grams                                                                                                                                                                         Average weight of pan 2, =          25        grams                                                                                          Percent Difference for =          0.8          %                                                                                                                                                                                   Percent Difference for =          0.8           % EXPERIMENT 204:  TORQUE: SECOND CONDITION OF EQUILIBRIUM

DATA and OBSERVATIONS

 TABLE 2.  Determining the Force needed to be in Equilibrium Trial    Computed Measured % difference 1    53 o 22.5   cm 7.5   cm 74.8   g 281   g 280   g 0.36    % 2    70o 22.5   cm 15   cm 74.8   g 119   g 120   g 0.84    %

 TABLE 3.  Determining the Weight of the Beam Trial    (Computed) (Measured) 1 13.7   cm 7.5   cm 74.8   g 136.6    g 136.6                  g 2 12.2   cm 7.5   cm 84.8   g 137.9   g 3 10.9   cm 7.5   cm 94.8   g 137.8   g Average Weight of Beam, =          137.4      grams Percent Difference =          0.58          %

GUIDE QUESTIONS

1.    Can two unequal weights be balanced in the model balance? How should the different weights be positioned in
the model balance to achieve equilibrium?

The two unequal weights can be balanced in the model balanced. The heavier weight must be positioned closer to the axis of rotation located at the center while the lighter weight must be positioned farther from the center than the position of the heavier weight.

2.    In part B of the experiment, how does the angle of the spring balance with the beam affect its reading? Does the angle affect the torque needed to balance the beam? Explain briefly.

The angle of the spring balance will not affect the torque needed to balance the beam. Applying a constant force, lets say, 3 N to the spring balance and an angle will only get the y component of that force and as the force is changing, the angle is also changing. This shows that the angle is the basis for a correct experiment. The angle measured must also agree with the force applied by the spring balance.

3.   In Part C of the experiment, is the torque due to the beam's weight constant for the three trials? How can the
torque due to the weight be changed?

The torque due to the beam's weight is constant for three trials. The torque due to the weight can be changed by changing the axis of rotation or the pivot point to a new distance. The torque due to the weight will change but the perpendicular force or the weight remains constant.

Problem:

1. A uniform meter stick with a mass of 250g is supported horizontally by two vertical strings, one at the 0.0-cm mark an the other at the 90.0-cm mark. (a.) What is the tension in each string? (b.) Where on the meter stick must a 150-g weight so that the tensions in the strings are the same?

ANALYSIS

When a weight is added in pan1, the position of pan1 must be closer to the axis of rotation than its original equilibrium position. Pan2 which has no additional weight must be farther than the distance of the lever arm of Pan1 and vise versa when weight is added on pan2.

In the part 2 of the experiment, the angle of the spring balance affects the equilibrium because the spring balance does not give an accurate reading when it is not hanged upright. What happened is that when it has an angle, the equilibrium can still be achieved but the resulting force measured will not be accurate.

In getting the weight of the beam, the support of the beam must be transferred to the second hole as the axis of rotation so that the center of gravity of the beam does not pass through the new axis of rotation. Equating the torque of the center of gravity of the beam and the torque of the pan and the added weight by letting the weight of the beam as the unknown will lead to the formula:
Wb= L1(W1+P1)
L2

CONCLUSION

1. Anonymous01:04

Bakit ba pinost pa to? -_-

1. Anonymous09:54

yabang

2. Anonymous13:59

Tanginang yan bat walang conclusion to?

1. Anonymous01:36

PUTANG INA MO KASI

3. Anonymous09:58

Putang ina conclusion! Burat!

1. Anonymous23:23

Mukha kang burat

4. Anonymous00:31

mga bobo. kaya kayo bumabagsak eh. reference lang to hindi dapat kinokopya mga tanga

1. Anonymous16:14

HAHAHHAHAHAHHAHA. may point ka eh.

5. Anonymous16:49

ULOL KAYONG LAHAT!

6. Anonymous01:11

sila na nakinabang, sila pa galit, edi aw.

7. Anonymous14:27

hahhahahaha kopya pa po =)))

8. Anonymous17:54

Mga gago

9. Anonymous18:02

gago ka din

10. Anonymous23:26

Mga trashtalker mga tanga