If there was slipping then the bike would stay put. The hub would be stationary and the wheel would be able to spin freely (as it slipped across the ground.) Since we are told that there is no slipping, when the point on the far end of the wheel moves toward the puller, the bike must move away as the wheel turns.
So the guy is standing or sitting. If he is sitting and pulling the string to the chest, it would still move forward.
But if he is standing and pulling the string, the bike might get some air and the wheel would rise up a little bit depending on the force of the pull and it would go forward for a sec or less and then fall over.
TuxDave
Lifer
- Oct 8, 2002
- 10,571
- 3
- 71
But it's like two points of rotation going in opposite directions. If p exists at a height between the axis and the ground (and not at either point), it will be exerting a torque for both cases and it will be in the opposite direction of rotation for each. Who's to say one automatically wins?
if there was slipping then the bike would move backwards because there would be no force of friction to move the boke forward.
force P isn't important, the moment exerted about the axle by force P is what matters. resolve force P into components parallel and perpendicular to the wheel at the point of contact. the parallel component goes down and to the right, exerting a counter clockwise moment. the bike goes forward.
here's anther way of looking at it, IF the bike moves forward, the point of contact with string would be moving left and down by vector addition. if moving backward, the point would move up and right. now tell me which makes more sense
Nope, look which way the force of friction is pushing. It is pushing left on the bottom of the wheel making it go clockwise. Wheel going clockwise = bike going backward. If there was partial slipping, the wheel would move counterclockwise, making the bike go forward.
actually the bike will move backwards. it would only stay put if it was on a completely frictionless surface
HardcoreRobot
Lifer
- Nov 7, 2000
- 16,403
- 3
- 81
not true, becuase the force is not purely rotational. imagine if the string was at the 180 degree position (halfway up on the left). there would be NO rotational force there, only a force to pull the entire wheel to the right. which, with slippage, would drag the wheel without spinning. the bike would still move. even though the string is not in this position, a component of its force acts like this.
edit: with friction, a force at that angle would not move the bike (would not rotate, nor would it slide)
It's NOT the moment exerted about the axel. The center of rotation is the point where the ground touches the wheel.
I think they are trying to see your analytical skills. I would answer something to the effect that it is impossible to know how the bike will move if we don't know the strength of the string. I was asked question like that in my old job because they wanted to see how my problem solving skills were.
ActuaryTm
Diamond Member
- Mar 30, 2003
- 6,858
- 12
- 81
Nope, the part that answers this is that there is no slip conditions between the wheel and the ground. The force does not matter because no matter how hard the string is pulled, the wheel will rotate about the ground at it's contact point. it will go backwards.
I'm majoring in mechanical engineering and have had a dynamics course or two. This question has come up in different forms.TRENDING THREADS
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