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I am stuck. I have a problem in my physics class. And i need to figure out torques. Say I am applying 80 N force and rotating a gear, the handle that im pushing on is 20 cm from center and the edge of the gear is 30 cm from the center. This gear rotates a smaller gear. What would torque be of the smaller gear if it has a 2 Nm torque in resistance to turning! Please help me! Another question that I have is - if I am turning this gear at a handle with 80 N force, what force would be at the edge of my gear? I appreciate it!

2006-10-10 11:31:20 · 2 answers · asked by Anonymous in Education & Reference Homework Help

smaller gear has a radius of 5 cm

2006-10-10 11:38:41 · update #1

2 answers

Note: smaller gear has a radius of 5 cm

Step 1.) The handle is a lever, so the force applied is F * ratio of handle length to gear radius (because the handle is connceted to the gear).
Fa = F * 20cm / 30cm = 2/3 * 80 = 53 1/3 N applied by the large gear on the small gear.

Step 2.) Remember: Torque = r * F.
F = 53 1/3N from above.
Total Torque = Torque applied - Torque resistance
Tt = 53 1/3 N * .05m - 2 N-m
Tt = 2 2/3 N - 2 N-m = 2/3 Nm torque on the small gear. (solution).

2.) I'm not sure which gear you're turning, so I'll assume it's the small gear - we already calculated the force at the edge of the large gear if the handle is attached to it.

If the handle is attached to the small gear, then the ratio of the handle length to gear radius is 4. The force is therefore 4 * force applied = 320 N.

The key here to remember is that a handle only is useful if it larger than the other side of the lever (be it a gear, or whatever) - otherwise the handle diminishes your force.

2006-10-11 02:59:06 · answer #1 · answered by ³√carthagebrujah 6 · 0 0

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2016-10-02 04:14:54 · answer #2 · answered by ? 4 · 0 0

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