Force on a Cyclist: Net Force Calculation

  • Thread starter Thread starter DarkEnergy890
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The discussion centers on the calculation of net force acting on a cyclist, where the net force is stated as 576N, derived from an applied force of 450N and a frictional force of 126N. Participants express confusion regarding the nature of forces, particularly the role of friction, which opposes motion and should be considered negative. It is clarified that the cyclist's force is not a net horizontal force, as internal forces between the cyclist and bicycle cancel out. The actual force propelling the bicycle comes from static friction, and the problem is criticized for being poorly posed and containing errors, particularly in the treatment of friction and the direction of forces. Overall, the conversation highlights the complexities of force calculations in cycling dynamics.
DarkEnergy890
Messages
14
Reaction score
3
Homework Statement
A cyclist exerts a constant force of 450 N causing the bicycle to accelerate at 4.8 m s-2
until reaches a velocity of 20 m s-1 due west. Given that the combined mass of the
cyclist plus the bicycle is 120 kg, calculate (i) the net force (ii) the frictional force.
Relevant Equations
F=ma
My understanding is that the Net Force = Force Applied + Frictional Force. The net force is F=ma, so net force = 576N. Now 576N=450N + Frictional force, so frictional force has to be 126N.

My confusion is this:
1. Force is a vector, and the frictional force opposes the direction of motion. Therefore shouldn't friction turn out to be a negative?
2. If the applied force is 450N, then how can the net force be 576N, more than the force applied? If anything, I think that friction should cause the net force to be less than the applied force.

Here is the marking scheme (model answer) : https://prnt.sc/tyhQTEm0WKAw

Thanks!
 
Physics news on Phys.org
DarkEnergy890 said:
My confusion is this:
1. Force is a vector, and the frictional force opposes the direction of motion. Therefore shouldn't friction turn out to be a negative?
The model answer clearly show the friction acts east, which is a direction. Directions can be north, south, east, west etc. Not just ##\pm##.
DarkEnergy890 said:
2. If the applied force is 450N, then how can the net force be 576N, more than the force applied? If anything, I think that friction should cause the net force to be less than the applied force.
A bicycle has pedals, gears and wheels that allow a force to be magnified, as it were. That's why it's possible to cycle uphill comfortably in low gear.
DarkEnergy890 said:
Here is the marking scheme (model answer) : https://prnt.sc/tyhQTEm0WKAw
That answer looks completely wrong to me. The force of the cyclist on the bicycle produces an equal and opposite force of the bicyle on the cyclist. These are therefore internal forces that cancel out.

The entire accelerating force must come from static friction of the tyres on the ground. Draw a free-body diagram and you'll see that the force from the ground on the bicycle must be ##576 \ N## (in the direction of the acceleration).
 
  • Like
Likes DarkEnergy890
PeroK said:
That's why it's possible to cycle uphill comfortably in low gear.
Speak for yourself! :wink:
 
  • Like
  • Haha
Likes Lnewqban and erobz
As @PeroK observes, the question is poorly posed and the given solution is nonsense.
I'll assume this is on the horizontal (which should have been stated).

By "friction" the author appears to mean rolling resistance. Confusing that with friction is a novice error.

Saying that the rider exerts a force of 450N is pretty meaningless. The rider cannot be exerting a net horizontal force on the bike or the two would part company. The vertical force, mg, is irrelevant. If it means the force exerted on one pedal (no cleats, then) we would need to know about the gearing etc.
To make any use of it to solve the questions asked, we have to assume it means that, in the absence of losses, the cyclist would cause there to be a propulsive force of 450N on the bike+rider system. The actual force would come from static friction on the tyres acting forwards.

Finally, there is the sign error involving the "friction". Having written the equation as ##F_{net}=F_{bike}-F_{friction}##, ##F_{friction}## must be being taken as positive to the East. Hence, getting -126N means the force is 126N West, aiding the acceleration!

Where does this sloppy problem come from?
 
  • Like
Likes DarkEnergy890, jbriggs444 and berkeman
haruspex said:
As @PeroK observes, the question is poorly posed and the given solution is nonsense.
I'll assume this is on the horizontal (which should have been stated).

By "friction" the author appears to mean rolling resistance. Confusing that with friction is a novice error.

Saying that the rider exerts a force of 450N is pretty meaningless. The rider cannot be exerting a net horizontal force on the bike or the two would part company. The vertical force, mg, is irrelevant. If it means the force exerted on one pedal (no cleats, then) we would need to know about the gearing etc.
To make any use of it to solve the questions asked, we have to assume it means that, in the absence of losses, the cyclist would cause there to be a propulsive force of 450N on the bike+rider system. The actual force would come from static friction on the tyres acting forwards.

Finally, there is the sign error involving the "friction". Having written the equation as ##F_{net}=F_{bike}-F_{friction}##, ##F_{friction}## must be being taken as positive to the East. Hence, getting -126N means the force is 126N West, aiding the acceleration!

Where does this sloppy problem come from?
Thanks for the detailed reply! This problem came from a "pre-paper" company - not an official exam, so that's probably why there are so many mistakes in it.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top