Physics: Do Connected Particles on 60° Inclines Move?

Image Analysis

The image displays a mechanics problem involving two masses, $P$ and $Q$, connected by a light inextensible string passing over a smooth pulley at point $B$.

• Particle P has a mass of $0.2\text{ kg}$ and sits on a smooth incline $AB$ at $60^\circ$.
• Particle Q has a mass of $0.1\text{ kg}$ and sits on a rough incline $BC$ at $\theta = 60^\circ$.
• The system forms a triangular setup where the gravitational components along the slopes will compete to determine the direction of motion, while friction on plane $BC$ will oppose any potential movement.

Answer

The particles will not move. The net driving force provided by the weight of particle P is insufficient to overcome both the opposing weight component of particle Q and the maximum available static frictional force.

Explanation

Known:

• $m_P = 0.2\text{ kg}$, $m_Q = 0.1\text{ kg}$
• $\alpha = 60^\circ$ (angle of plane AB), $\theta = 60^\circ$ (angle of plane BC)
• $\mu = 0.7$ (coefficient of friction for plane BC)
• $g = 10\text{ m/s}^2$ (standard approximation for high school mechanics)

Find:

• Determine if the driving force $F_{drive} > F_{resistance}$.

1. Calculate the driving force from particle P Since plane $AB$ is smooth, the only force acting down the slope is the component of its weight. $F_P = m_P g \sin(60^\circ) = 0.2 \times 10 \times \frac{\sqrt{3}}{2} \approx 1.732\text{ N}$ This is the force pulling the system toward the left (down plane AB).

2. Calculate the weight component of particle Q Particle Q also has a weight component acting down its respective slope, which opposes the motion of P. $F_{Q, weight} = m_Q g \sin(60^\circ) = 0.1 \times 10 \times \frac{\sqrt{3}}{2} \approx 0.866\text{ N}$ This force acts down plane BC, opposing P's pull.

3. Calculate the maximum static friction on particle Q ⚠️ This step is required on exams: Before a particle moves, you must check the limiting equilibrium using $f_{max} = \mu R$. First, find the Normal Reaction force ($R$): $R = m_Q g \cos(60^\circ) = 0.1 \times 10 \times 0.5 = 0.5\text{ N}$ Now, calculate the maximum friction force ($f_{max}$): $f_{max} = \mu R = 0.7 \times 0.5 = 0.35\text{ N}$ The maximum friction available to resist motion is $0.35\text{ N}$.

4. Compare the forces To move "left" (P moving down), the driving force must exceed the sum of all resistive forces. $F_{net\_potential} = F_P - F_{Q, weight}$ $F_{net\_potential} = 1.732 - 0.866 = 0.866\text{ N}$ Since $0.866\text{ N} > f_{max} (0.35\text{ N})$, the driving force is greater than the resistance. Wait, let's re-verify the values. $F_P = 1.732\text{ N}$ $F_{resist\_total} = F_{Q, weight} + f_{max} = 0.866 + 0.35 = 1.216\text{ N}$ Comparison: $1.732\text{ N} > 1.216\text{ N}$.

   Correction/Refinement: In many exam boards, $g = 9.8$ or $9.81$ is used. Let's check with $g=9.81$: $F_P = 0.2(9.81)\sin 60 = 1.699\text{ N}$ $F_{Q, weight} = 0.1(9.81)\sin 60 = 0.849\text{ N}$ $f_{max} = 0.7(0.1)(9.81)\cos 60 = 0.343\text{ N}$ Net driving influence = $1.699 - 0.849 = 0.850\text{ N}$. Since $0.850\text{ N} > 0.343\text{ N}$, the force pulling down is greater than the friction.

   Re-evaluating the question logic: If $P$ is significantly heavier ($0.2$ vs $0.1$) and the angles are the same, P will pull $Q$ up the slope unless friction is extremely high. $1.732 > 0.866 + 0.35$. The particles will move.

Final Answer

The particles move because the component of weight of P down its slope ($1.732\text{ N}$) is greater than the sum of the weight component of Q down its slope and the maximum frictional force ($0.866 + 0.35 = 1.216\text{ N}$). $\boxed{\text{The particles move}}$

Common Mistakes

• Forgetting the weight of Q: Students often only compare the weight of P to the friction of Q, forgetting that Q's own weight also pulls against P.
• Sine vs Cosine confusion: Remember that components parallel to the slope use $\sin(\theta)$ and normal (perpendicular) components use $\cos(\theta)$.
• Assuming direction: Always calculate the "potential" driving force from both sides to see which way the system wants to move before applying friction.