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1. On a see-saw mass A is 60 kg
and is 1 m from the pivot. Mass B is 30 kg
and mass C is 10 kg. Mass C is 3 m from
the pivot. The see saw is at an angle of 30
degrees from the horizontal. Calculate the
torques of A and C about the pivot and
where you should place mass B for the
torques to balance.



2. As part of a physical therapy program
following a knee operation, a 5-kg object is
attached to an ankle and leg lifts are done as
sketched in the figure. Calculate the torque
about the knee due to this weight for the four
positions shown. from the knee to the weight is 40 cm
the four positions are at 0, 30, 60, 90 degrees.

2007-12-06 19:46:20 · 2 answers · asked by bebop1223@sbcglobal.net 2 in Science & Mathematics Physics

2 answers

If you draw a diagram of the see-saw, and the angle is sloping up towards mass C, then the torque from mass A is -60kg*1m*g*cos(30º) and the torque from mass C is 10kg*3m*g*cos(30º). For equilibrium all torques must sum to zero, so add the torque from A to the torque from C (observing the sign difference). If that sum is not zero, the torque from B must equal the negative of the remainder. (Since g and cos30º appears in all terms, we can compute the balance without them.)

∆T = 10kg*3m -60kg*1m = -30

r*30 = 30, r = 1m; the weight is then on the same side as weight C.

For the second part, remember that torque is the force (mass times g for weights) times the distance along a line drawn perpendicular to the direction of the force and through the pivot.

Note regarding the angle: If the angle between the force vector and vector from axis to point where force is applied is ø, then the torque is F*r*sinø. However, in this problem, the angle is given as 30º from the *horizontal*, but the force (weight) is *vertical*, so the angle is 90º - ø, and the torque becomes F*r*cosø.

2007-12-06 20:29:29 · answer #1 · answered by gp4rts 7 · 1 0

1) moment = Force X perpendicular distance from the pivot

moments of A and c .

moments of A = 60(9.8)(1)sin30=294 Nm
moments of C= 10(9.8)(3sin30) =147 Nm
294-147 = 147 Nm .

Now to balance be must provide moment of 147 Nm
30(9.8)xsin30 = 147 .
x=1 m from pivot .

2)Can you send me the figure .
upload it on www.photobucket.com or mail me .

2007-12-07 04:42:16 · answer #2 · answered by Murtaza 6 · 0 0

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