The figure below is a long, massless stick with evenly spaced marks. The distanc

Question

The figure below is a long, massless stick with evenly spaced marks.

The distance between two adjacent marks is 5.50 cm. In each of the following problems a pivot will be placed at a point on the stick. The pivot will hold that point fixed, but allow the stick to rotate. You will be finding conditions that will keep the stick in static equilibrium given different forces. All forces will be applied to the stick perpendicular to the length of the stick.

1) A pivot is placed at Point K and a force of 22.0 N is directed downward perpendicular to the stick at Point J. What is the magnitude and direction of the force which must be applied at point G in order to obtain static equilibrium?
Upward or Downward?

2) A pivot is placed at Point G and a force of 34.0 N is directed downward perpendicular to the stick at Point L. What is the magnitude and direction of the force which must be applied at point B in order to obtain static equilibrium?
Upward or Downward?

3) A pivot is placed at Point D and a force of 33.0 N is directed downward perpendicular to the stick at Point F. One at a time, each of the following 3 forces are applied to the stick in the manner described. Where must these forces be applied in order to achieve static equilibrium? If the is no point that will work within reason, select the option None.

-A force of 9.9 N directed upward perpendicular to the stick?

A, B, C, D, E, F, G, H, I, J, K, L, None

-A force of 19.8 N directed upward perpendicular to the stick.

A, B, C, D, E, F, G, H, I, J, K, L, None

-A force of 19.8 N directed downward perpendicular to the stick.

A, B, C, D, E, F, G, H, I, J, K, L, None

Explanation / Answer

1) One side of the fulcrum is zero, the other side we have 22N*5.5cm-22cmX. So X= 5.5N upwards beacuse it needs to go against the other force of 22N.

2)Since Point L and point B are both the same distance from the fulcrum, the same force is needed. So point B would be 34N downward, because it is on the other side of the fulcrum.

3)A) between the J and K, because the force is upwards we know it must be on the same side. So again we have 0=33*(5.5*2)-(9.9*5.5*X) thus X is 6.6 so it is 6.6 spots away from D.

B)Between the G and the H, It is the same concept as part A.

C) Similar to part B, although now the force is downwards, so this time it's 3.3 spaces away from the fulcrum on the left side, although, the answer to this is none, because it is unrealistic because A is the end of the stick

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