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A few notes about Assignment 2:

Problem 1

The main confusion for people was with the free body diagram. This is a full-strength FBD problem with forces in all directions and moments about all axes. It may help you to sketch in the gear on the counter shaft that the input gear mates with. This gear is just below the input gear. There are contact forces in all 3 directions (XYZ) at the contact because they are helical gears. Furthermore, these contact forces are not in the same vertical plane as the main ball bearing. So they produce moments about the bearing, which must be balanced by reaction forces.

The little roller bearing inside the the end of the input shaft can only support radial forces. Its main job is to stabilize the end of the rather long output shaft that is cantilevered into the transmission and would otherwise be a bit wobbly.

Problem 2

For the car, a few people had questions about how to do the free body diagram. Probably the easiest way to do problems like this is to recall that a wheel rolling on the ground is also instantaneously rotating about its contact point with the ground, with an angular velocity of r_w * omega_w, where r_w is the wheel radius and omega_w is the wheel angular velocity. So we can put a coordinate system at the center of the wheel-ground contact patch and balance X, Y forces and moments there. Attached below are a couple of diagrams from related problems in recent years. As you can see, there is more than one reasonable way to set it up. The essential thing is to be consistent about how you take your forces and moments:

  • Attach:ProblemTwoDiagrams.pdf -- Two logical ways to think about the balance of input torque versus drag and rolling resistance. They give the same end result but version 'A' probably gives a better estimate of the required traction force.
  • Attach:wheel-roll.pdf -- A similar diagram for a car (or Lego crawler) climbing a hill. Replace the air drag with mg sin(theta) and you have something almost the same as in Problem 2. At the bottom of the notes the words "P_in" and "P_climb" and "P_loss" have been partly clipped off. This diagram shows how T_roll can be obtained in a detailed way as the result of the normal reaction force being not quite at the center of the contact patch and therefore producing a (small) moment about it.

FAQ and responses

For Problem 2.1, are F_roll and F_r the same thing? It states that they're independent of one another, but if they are indeed different than I'm not quite sure how to solve for mu to complete 2.2.

This is a key point and it will come up in the crawler projects too, which is part of why we wanted to give a preview of it on Assignment 2, to get people used to the idea.

F_r is the tangential component of the reaction force between the wheel and the ground. It is necessary for traction and propulsion. It is limited by F_r < mu * F_n due to Coulomb friction. If you are driving on ice, this is a problem because mu is very low, so F_r is very low. But even on a perfectly icy road, you could still have rolling resistance. And note that you don't actually need to know mu to complete the problem. It's just a limit to be aware of.

F_roll, or equivalently T_roll = F_roll * r_wheel, is a different thing. It is rolling resistance due to the flexing of the rubber tires as they roll. A bicycle with fat knobby tires has a larger T_roll than a racing bike. But if you are going up a hill, the traction F_r is the same for both bikes because in both cases it has to balance the gravity effect mg sin(theta).

Certain formulations of rubber can be used that have less rolling resistance, which is why cars like the electric Nissan Leaf and the plugin Prius have special low-rolling-resistance tires. Note that although they have low rolling resistance, they still have good traction (Fr) -- otherwise they would do poorly in braking in an emergency! Unfortunately, however, these tires are also more expensive and sometimes don't last quite as long as ordinary tires (as my wife just discovered).

Page last modified on January 16, 2014, at 09:41 PM