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course phy 201
I'm not sure if I already submitted this
031. Torques and their effect on rotational motion
Question: `q001. Note that this assignment contains 9 questions.
Imagine that you are turning the top on a jar of peanut butter. The top is on pretty tight and you have to use the fair amount strength to get the top loose. You can squeeze the top as tightly as you like, but unless you also turn the top it is not going to come loose. However you do know from experience that you do have to squeeze it pretty tightly, or your hand will just slide around the top instead of turning it.
The reason you have to squeeze and turn is that you use the frictional force between your hand and the top of the jar to transmit the turning force exerted by your arm muscles.
The squeezing force is directed toward the center of the circular top and is therefore perpendicular to the arc of the top. It has no rotational effect. The frictional force, by contrast, is directed along the sides of the jar's top, at every point parallel to the arc of the circle and hence perpendicular to a radial line (a radial line is a line from the center of the jar to a point on the circle; the radial line in this case runs from the center to the point at which the frictional force is applied).
This type of force causes a turning effect on the top, called a torque.
The amount of the torque depends on how much force is exerted parallel to the arc of the circle, as well as on how far the force is exerted from the center of rotation. For example, if you exert a force of 50 Newtons in the direction of the sides, on a top of radius 4 centimeters, the torque would be 50 Newtons * 4 cm = 200 cm * Newtons.
If the cap is too tight, you might use a pipewrench to turn it. The pipewrench 'grabs' the top and allows you to exert your force at a point further from the center of the top. You naturally push in a direction perpendicular to the handle of the wrench, which is pretty much perpendicular to the line from the center to the point at which you push. So for example you might exert a force of 20 Newtons at a distance of 15 cm from the center of the top, resulting in a torque of 20 N * 15 cm = 300 cm * N. This torque, though it results from less force, is greater than the torque exerted in the previous calculation.
What would be greater, the torque exerted by a 70 Newton force at a distance of 4 cm from the center of the top, or the torque exerted by a 20 Newton force at a distance of 15 cm from the center of the top?
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Your solution:
70 * 4 = 280
20 * 15 = 300 the 20 newtons is greater
confidence rating #$&*:
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Given Solution:
The first force would be 70 Newtons * 4 cm = 280 cm N, while the second would be 20 N * 15 cm = 300 cm N, so the second would be the greater. This second torque would be more likely to succeed in opening the jar.
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