Question

A bicycle rider pedals up a hill. The combined mass of the rider and the bicycle is $M$. Which of the followings is/are correct?

• A. The friction force must be directed down the slope of the hill.
• B. The net force applied by the hill on the bicycle must be greater than $Mg$ if the bicycle is moving with increasing speed.
• C. The net force applied by the hill on the bicycle may be equal to $Mg$.
• D. The bicycle may retard even if friction is directed up the slope of the hill.

Answer 1

Answer: (B), (C), (D)

(b), (c), (d)

Answer 2

The problem asks us to analyze the forces acting on a bicycle rider pedaling up a hill and determine the correctness of several statements. Let $M$ be the combined mass of the rider and the bicycle, and $\theta$ be the angle of inclination of the hill. We set up a coordinate system with the x-axis along the slope (positive upwards) and the y-axis perpendicular to the slope (positive outwards).

The forces acting on the bicycle are:

1. Gravitational force ($Mg$): Acts vertically downwards. Its components are $Mg \sin\theta$ down the slope and $Mg \cos\theta$ perpendicular to the slope, into the hill.
2. Normal force ($N$): Exerted by the hill, perpendicular to the slope, outwards.
3. Friction force ($f$): Exerted by the hill, parallel to the slope. For a bicycle to move up, the rear wheel pushes backward on the ground, so the ground pushes forward on the wheel. This forward push is the static friction force, which acts up the slope for propulsion.
4. Pedaling force ($F_p$): This is an internal force within the rider-bicycle system. The effect of pedaling is to generate a torque on the rear wheel, which in turn leads to the friction force $f$ from the ground. So, $f$ is the actual propulsive force from the external environment.

From Newton's second law:

• Perpendicular to the slope: $N - Mg \cos\theta = 0 \implies N = Mg \cos\theta$.
• Along the slope: $f - Mg \sin\theta = Ma$, where $a$ is the acceleration along the slope. (Here, $f$ represents the net friction force providing propulsion or opposing motion).

Let's evaluate each option:

(a) The friction force must be directed down the slope of the hill.

When a bicycle rider pedals up a hill, the driving wheel (usually the rear wheel) pushes backward on the ground. By Newton's third law, the ground pushes forward on the wheel. This forward force is the static friction force, which acts up the slope to propel the bicycle forward. Therefore, this statement is incorrect.

(b) The net force applied by the hill on the bicycle must be greater than $Mg$ if the bicycle is moving with increasing speed.

The net force applied by the hill on the bicycle is the vector sum of the normal force $\vec{N}$ and the friction force $\vec{f}$. Let its magnitude be $F_{hill}$.

$F_{hill} = \sqrt{N^2 + f^2} = \sqrt{(Mg \cos\theta)^2 + f^2}$.

If the bicycle is moving with increasing speed up the hill, its acceleration $a$ must be positive ($a > 0$).

From the equation along the slope: $f - Mg \sin\theta = Ma$.

Since $Ma > 0$, it implies $f - Mg \sin\theta > 0$, or $f > Mg \sin\theta$.

Now, let's compare $F_{hill}$ with $Mg$:

Is $F_{hill} > Mg$?

$\sqrt{(Mg \cos\theta)^2 + f^2} > Mg$

Squaring both sides: $(Mg \cos\theta)^2 + f^2 > (Mg)^2$

$M^2 g^2 \cos^2\theta + f^2 > M^2 g^2$

$f^2 > M^2 g^2 (1 - \cos^2\theta)$

$f^2 > M^2 g^2 \sin^2\theta$

Taking the square root (and since $f$ is positive): $f > Mg \sin\theta$.

Since we established that $f > Mg \sin\theta$ for increasing speed up the hill, this condition is met.

Therefore, the net force applied by the hill on the bicycle must be greater than $Mg$. This statement is correct.

(c) The net force applied by the hill on the bicycle may be equal to $Mg$.

For $F_{hill}$ to be equal to $Mg$:

$\sqrt{(Mg \cos\theta)^2 + f^2} = Mg$

$(Mg \cos\theta)^2 + f^2 = (Mg)^2$

$f^2 = (Mg)^2 - (Mg \cos\theta)^2 = (Mg)^2 (1 - \cos^2\theta) = (Mg)^2 \sin^2\theta$

So, $f = Mg \sin\theta$.

From the equation of motion along the slope: $f - Mg \sin\theta = Ma$.

If $f = Mg \sin\theta$, then $Mg \sin\theta - Mg \sin\theta = Ma \implies Ma = 0$.

This means $a = 0$.

So, if the bicycle is moving at a constant speed (or is at rest) up the hill, the acceleration is zero, and the friction force required is exactly $Mg \sin\theta$. In this scenario, the net force applied by the hill on the bicycle would be exactly $Mg$. This statement is correct.

(d) The bicycle may retard even if friction is directed up the slope of the hill.

Retardation means $a < 0$. Friction directed up the slope means $f > 0$.

From the equation of motion along the slope: $f - Mg \sin\theta = Ma$.

If $a < 0$, then $Ma < 0$.

So, $f - Mg \sin\theta < 0 \implies f < Mg \sin\theta$.

This scenario is physically possible. For example, if the rider is pedaling very lightly, or has stopped pedaling but not applied brakes, while moving up the hill. The component of gravity down the slope ($Mg \sin\theta$) might be greater than the friction force ($f$) acting up the slope (which could still be positive, providing some lift or preventing slipping). In this case, the net force would be down the slope, causing deceleration (retardation) while the bicycle is still moving up the hill.

Therefore, the bicycle may retard even if friction is directed up the slope of the hill. This statement is correct.

Based on the analysis, options (b), (c), and (d) are correct.