Register to reply 
Bicycle over curb 
Share this thread: 
#1
Mar1107, 05:49 AM

P: 80

1. The problem statement, all variables and given/known data
You are trying to raise a bicycle wheel of mass m and radius R up over a curb of height h. To do this, you apply a horizontal force F. What is the least magnitude of the force F_vec that will succeed in raising the wheel onto the curb when the force is applied at the center of the wheel? 2. Relevant equations F = 0 torque_total = 0 3. The attempt at a solution I've tried for the last hours to figure out how to do this but I'm stuck. I've drawn a free body diagram with the horizontal force and also, the force the curb is exerting on the wheel as well? They must equal right..? I don't really have any idea how to start this problem other than that though. So frusterated sigh. :( Any help is appreciated, thanks. Pic: http://img85.imageshack.us/img85/945...ure1142xt0.jpg 


#2
Mar1107, 08:00 AM

Mentor
P: 41,312

First identify all the forces exerted on the wheel. Then analyze the torques they exert about the pivot point. What's the pivot point as the wheel begins to climb the curb?



#3
Mar1107, 12:11 PM

P: 312

If youve already drawn a free body diagram, draw a torque diagram. At what distance does each force act over?



#4
Mar1107, 04:19 PM

P: 80

Bicycle over curb
Thanks for the help.
I'm stuck on the forces. So far, I've identified gravity force, normal force from the ground, the force the curb is exerting on the wheel and the horizontal force. This is all right? But normal force and gravity force shouldn't affect the motion because they're in the y direction right? So it's only the horizontal force and curb force that matter....correct me if I'm wrong. As for the pivot point and the distance each force act over, I'm confused there also. I think it should be where the wheel and the curb meet? But I'm not sure because that would make it very hard to find the torque for the force the curb is exerting on the wheel..? Unless there's something I'm missing... 


#5
Mar1107, 05:46 PM

Mentor
P: 41,312




#6
Mar1107, 06:35 PM

P: 80

The normal force should be zero because there's nothing pushing up anymore. For the torques, this is what I have so far:
Gravity  I'm confused. I'm thinking it should be mgR, because R is the distance from the gravity force (which should be at the center of mass right?) to the pivot point. I don't understand what you mean by perpendicular distance... Normal force  0 F of the curb  0 Horizontal force  FR 


#7
Mar1107, 06:42 PM

Mentor
P: 41,312

Torque is not just force*distance or FDthe angle the force makes makes a difference! One way to express torque is Force*perpendicular distance (also called "moment arm"); another way is Force*distance*sin(theta).
Look up the definition of torque in your text. Read this: Torque 


#8
Mar1107, 07:14 PM

P: 80

Ah erm, so here I go again.
Horizontal force  The perpendicular distance of F to the pivot line, is it (R  h)? So then that would make the torque = F(R  h)? I'm not sure about theta because it's not given in the problem. Unless there's some other logic that I'm missing again. Gravity  It should still be mgR right? Because the perpendicular distance is just R and gravity force points downward so... Am I wrong again? 


#9
Mar1107, 07:37 PM

Mentor
P: 41,312




#10
Mar1107, 08:00 PM

P: 80

Thanks for all your help, really appreciate it. I got it now (cleared the torque confusion too). Phew.



Register to reply 
Related Discussions  
Parking uphill without a curb  question  General Physics  11  
'Car' vs curb  Introductory Physics Homework  2  
Effective ways to curb vandalism?  General Discussion  22  
You all need to Curb Your Enthusiasm  General Discussion  17 