Free body diagram - cycling

[chiddler]

When rapidly turning a corner on a flat road, a cyclist leans to the center of the turn. The frame of the bike is nearly parallel to which vector?


C. The centripetal force (67%, 36 Votes)
B. The normal force on the pair (17%, 9 Votes)
D. The sum of the normal and friction forces (15%, 8 Votes)
A. The force of gravity on the bicycle and rider (1%, 1 Votes)

Answer is D!

Normal force is perpendicular to surface, always. Friction is backwards relative to the front of the bike so combining those two vectors: Up + backwards does not equal the leaning bike frame!

Not sure if i'm right, but i'm not sure if the supposed answer is correct either.

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[Ssina]

i think I would go for D also, mostly through POE.. with that said, the force of friction is static and I think it's pointing forward, not backwards.

 

[Pisiform]

none of them makes sense to me lol

 

[Ssina]

I just know it should'be be A, B, D... 3 wrong enough to make a right on MCAT lool
with that said, OP do you have the answer?!

 

[MedPR]

When rapidly turning a corner on a flat road, a cyclist leans to the center of the turn. The frame of the bike is nearly parallel to which vector?


C. The centripetal force (67%, 36 Votes)
B. The normal force on the pair (17%, 9 Votes)
D. The sum of the normal and friction forces (15%, 8 Votes)
A. The force of gravity on the bicycle and rider (1%, 1 Votes)

Answer is D!

Normal force is perpendicular to surface, always. Friction is backwards relative to the front of the bike so combining those two vectors: Up + backwards does not equal the leaning bike frame!

Not sure if i'm right, but i'm not sure if the supposed answer is correct either.

If you are on flat ground, then the normal force and the force of gravity are in the same dierction. B and A are both wrong, since they are equivalent.

The sum of the normal force and the frictional force, assuming you mean the frictional force opposing the centripetal force is the mirror image of the frame of the bike. As in, the frame of the bike points up and left, while the sum of the normal force and frictional force points up and right.

The centripetal force is definitely not parallel..

Are any of the answers right?

 

[Pisiform]

If you are on flat ground, then the normal force and the force of gravity are in the same dierction. B and A are both wrong, since they are equivalent.

The sum of the normal force and the frictional force, assuming you mean the frictional force opposing the centripetal force is the mirror image of the frame of the bike. As in, the frame of the bike points up and left, while the sum of the normal force and frictional force points up and right.

The centripetal force is definitely not parallel..

Are any of the answers right?

lol yep ... thats what I thought

 

[MedPR]

lol yep ... thats what I thought

Wait a minute. The normal force points up and the frictional force points left.

If you line up the head of the frictional force vector with the tail of the normal force vector, then draw the resultant vector from the tail of the frictional force to the head of the normal force, you have a vector parallel to the frame of the bike!

I had my frictional force vector pointing the wrong way at first. The frictional force is the force preventing the tire from slipping (thus, the frictional force is preventing the rider from falling over).

Normal force vector plus <--------- vector = up and left vector.

 

[Pisiform]

ahh I see ... good question though

 

[chiddler]

I have two questions:

1. Unrelated, but for understanding: If the biker was not leaning, then static friction force would be pointing behind him, right? Because as he pedals there is torque and I remember that image that was posted with the last similar question i had asked.

2. Why is friction force pointed to the biker's right as he leans in to the right?

 

[MedPR]

I have two questions:

1. Unrelated, but for understanding: If the biker was not leaning, then static friction force would be pointing behind him, right? Because as he pedals there is torque and I remember that image that was posted with the last similar question i had asked.

2. Why is friction force pointed to the biker's right as he leans in to the right?

1. I'm pretty sure the frictional force of someone pedaling upright (same as driving a car) would be in the direction of motion. Your tire pushes backwards as you move forward, and the road opposes the force that your tire puts on it (equal and opposite)

2. Try pushing on a table with your finger at an angle. If the table is very smooth (no friction) your finger will easily slide along the table in the horizontal direction you are pushing in. If there is a lot of friction (sticky table) your finger won't move or won't move as freely. The reason why your finger doesn't move = friction = same as why the friction on a the leaning biker is in the same direction that he leans.

 

[MT Headed]

I have two questions:

1. Unrelated, but for understanding: If the biker was not leaning, then static friction force would be pointing behind him, right? Because as he pedals there is torque and I remember that image that was posted with the last similar question i had asked.

2. Why is friction force pointed to the biker's right as he leans in to the right?

(1) well, the frictional force on the tire would point forward, but you are changing the scenario entirely by having the biker push against the ground (via the tires). The original question didn't really have the biker pedaling, but rather leaning into a turn at constant coasting speed, and it was friction that held the biker up. With out friction (i.e. on ice) the biker would lean and the bike would slip & fall to the ground.

(2a) because frictional force always opposes a slipping motion. In the absence of friction, a biker leans to the right and the tires slip out to the left. Friction opposes this motion.

(2b) because if a biker leans to the right, we know from common sense that the bicycle will turn to the right, in a circle. Thus the Fnet on the biker must point inward towards the center of this circle. What provides this force? Gravity is down. Normal force is up. Friction must be supplying the Fnet inwards so the bicycle will go in a circle.

 

[chiddler]

2. Try pushing on a table with your finger at an angle. If the table is very smooth (no friction) your finger will easily slide along the table in the horizontal direction you are pushing in. If there is a lot of friction (sticky table) your finger won't move or won't move as freely. The reason why your finger doesn't move = friction = same as why the friction on a the leaning biker is in the same direction that he leans.

Ok! Spent a long time thinking about this. Understand. Thanks!


@MT Headed

2(a), very well put. That helped a lot. 2(b), very nice way to look at it! I never thought of it that way.

Does this mean friction applies in two direction: One opposing the forward movement of the bike and the other allowing the lateral motion?

 

[MedPR]

Ok! Spent a long time thinking about this. Understand. Thanks!


@MT Headed

2(a), very well put. That helped a lot. 2(b), very nice way to look at it! I never thought of it that way.

Does this mean friction applies in two direction: One opposing the forward movement of the bike and the other allowing the lateral motion?

Only if the bike is slanted and moving in a perpendicular direction (like turning). I don't think there is lateral friction on an upright bike; at least not enough to affect calculations.

Whenever I have a friction problem, I try to remember that for every force, there is an equal and opposite force. So as you slide down an incline, the friction pushes you back up the incline (not really, but you get it) and as you walk on concrete, everytime you push off (backwards) to step forward, the frictional force between the concrete and your shoe pushes your foot forward. Equal and opposite!

 

[chiddler]

Only if the bike is slanted and moving in a perpendicular direction (like turning). I don't think there is lateral friction on an upright bike; at least not enough to affect calculations.

Whenever I have a friction problem, I try to remember that for every force, there is an equal and opposite force. So as you slide down an incline, the friction pushes you back up the incline (not really, but you get it) and as you walk on concrete, everytime you push off (backwards) to step forward, the frictional force between the concrete and your shoe pushes your foot forward. Equal and opposite!

😀

it all makes sense now!