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Torque Problems

Torque Problems. In the following figure, a force whose magnitude is 55 N is applied to a door. However, the lever arms are different in the three parts of the drawing: a) l =.8 m, b) l =.6 m, and c) l =0m. Determine the torque in each case.

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Torque Problems

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  1. Torque Problems

  2. In the following figure, a force whose magnitude is 55 N is applied to a door. However, the lever arms are different in the three parts of the drawing: a) l =.8 m, b) l =.6 m, and c) l =0m. Determine the torque in each case.

  3. The figure shows the ankle joint and the Achilles tendon attached to the heel at point P. The tendon exerts a force of magnitude F=720 N. Determine the torque of this force about the ankle joint, which is located 3.6 x 10-2 m away from point P.

  4. Tracy is building a mobile to hang over her baby's crib. She hangs a 0.02 kg toy sailboat 0.01 m from the left end and a 0.015 kg toy truck 0.2 m from the right end of a bar 0.5 m long. If the lever arm itself has negligible mass, where must the support string be placed so that the arm balances?

  5. Your first step is to determine how much force there is in the support string. This is arrived at by getting the summation of all the vertical forces and since the system has to be in equilibriumForce in the support string (call this Fs) = 0.015(9.8) + 0.02(9.8)= 0.035(9.8) Fs

  6. The next step is to locate where Fs will be along the mobile and this is done by getting the summation of all torques and again, since this system is in equilibrium, the summation of torques = 0.

  7. The next step is to locate where Fs will be along the mobile and this is done by getting the summation of all torques and again, since this system is in equilibrium, the summation of torques = 0. • Let X = distance from the right end of the bar where the support string will be placed, in other words, the far right can be the axis of rotation • Watch signs

  8. -(9.8(0.015)(0.2)) + 9.8(0.035(X)) - (9.8(0.02)(0.49)) = 0 X = .366 m from the right side

  9. Figure 1 shows an object with several forces acting on it. The pivot point is at O. F1 = 10 N, and is at a distance of 0.25 m from O, where  = 80oF2 = 7.0 N, acting perpendicular to the object, at a distance of 1.25 m from O F3 = 12 N, is 0.60 m from O, and acts at  = 40o from the horizontal Find the total (net) torque on the object.

  10. = -10(sin80)(.25) + 12(sin40)(.6) – 7(1.25) = -6.58 Nm (where – is down)

  11. You spin a bicycle wheel (diameter of 0.85 m, mass of 4.5 kg), applying a force of 24 N tangentially. Find the angular acceleration of the wheel, assuming the spokes are negligible.

  12.  = I Fl = I 24 x .425 = MR2  24 x .425 = 4.5 x .4252 x   = 12.55 rad/s2

  13. More help

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