This problem involves two masses connected by a string passing over a frictionless pulley, with one mass on an incline. We need to find the coefficient of kinetic friction $\mu_k$ and verify the coefficient of static friction $\mu_s$.

Given:

• $m_1 = 9.24 \, \text{kg}$
• $m_2 = 7.70 \, \text{kg}$
• Incline angle $\phi = 54^\circ$
• Acceleration $a = 1.391 \, \text{m/s}^2$
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$

Step 1: Determine the forces on each mass

For $m_1$ (hanging mass):

The forces acting on $m_1$ are:

• Weight: $m_1 g$
• Tension in the string: $T$

The equation of motion for $m_1$, accelerating downwards: $m_1 g - T = m_1 a$

For $m_2$ (on the incline):

The forces acting on $m_2$ are:

• Weight: $m_2 g$
• Normal force: $N$
• Tension in the string: $T$
• Kinetic friction force: $f_k = \mu_k N$

The components of the forces:

• The gravitational force acting parallel to the incline: $m_2 g \sin \phi$
• The gravitational force acting perpendicular to the incline: $m_2 g \cos \phi$

The equation of motion for $m_2$, accelerating upwards (along the incline): $T - m_2 g \sin \phi - f_k = m_2 a$ Substitute $f_k = \mu_k N$, where $N = m_2 g \cos \phi$: $T - m_2 g \sin \phi - \mu_k m_2 g \cos \phi = m_2 a$

Step 2: Solve for $T$

From the equation for $m_1$: $T = m_1 g - m_1 a$ Substitute the given values: $T = 9.24 \times 9.8 - 9.24 \times 1.391 = 90.552 - 12.843 \approx 77.709 \, \text{N}$

Step 3: Solve for $\mu_k$

Substitute $T$ into the equation for $m_2$: $77.709 - 7.70 \times 9.8 \times \sin 54^\circ - \mu_k \times 7.70 \times 9.8 \times \cos 54^\circ = 7.70 \times 1.391$ First, calculate the trigonometric and other values:

• $\sin 54^\circ \approx 0.809$
• $\cos 54^\circ \approx 0.587$
• $7.70 \times 9.8 \times \sin 54^\circ \approx 61.071$
• $7.70 \times 9.8 \times \cos 54^\circ \approx 44.413$
• $7.70 \times 1.391 \approx 10.711$

Now substitute these into the equation: $77.709 - 61.071 - \mu_k \times 44.413 = 10.711$ Simplify: $16.638 - \mu_k \times 44.413 = 10.711$ $\mu_k \times 44.413 = 16.638 - 10.711 = 5.927$ $\mu_k = \frac{5.927}{44.413} \approx 0.134$

Step 4: Solve for the coefficient of static friction ($\mu_s$)

To prevent $m_2$ from sliding initially, the static friction must balance the forces along the incline. The maximum static friction force is given by: $f_s = \mu_s N = \mu_s m_2 g \cos \phi$ Setting the maximum static friction equal to the parallel component of gravity: $\mu_s m_2 g \cos \phi = m_2 g \sin \phi$ $\mu_s = \frac{\sin \phi}{\cos \phi} = \tan \phi$ $\mu_s = \tan 54^\circ \approx 1.376$

Thus, the coefficient of static friction must be less than 1.376 to allow sliding to begin.

Final Answers:

• The coefficient of kinetic friction $\mu_k$ is approximately 0.134.
• The coefficient of static friction $\mu_s$ is less than 1.376.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How would increasing the angle of the incline affect the required kinetic friction coefficient?
2. What happens if the frictionless pulley had friction?
3. How would the system behave if $m_1$ was heavier than $m_2$?
4. How is the normal force on $m_2$ related to the incline angle?
5. How would the motion change if the pulley were not frictionless?

Tip: Always break down forces into components parallel and perpendicular to the incline when solving such problems.