Atwood at an incline accelerating down

  • #1
Enginearingmylimit
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Homework Statement
A system comprising blocks, a light frictionless pulley, a frictionless, incline, and connecting (“massless”) ropes is shown in the figure. The 9 kg block accelerates downward when the system is released from rest. What is the tension in the rope connecting the 6 kg and 4 kg block?
Relevant Equations
F = ma
Fgy = 9.8 × m
Both myself and my TA gave up, but we found acceleration of the system

9g - 4gsin(30) = 13a
a=5.27m/s^2
 

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  • #2
Yes, you do need to find the common acceleration of the masses first but you have the wrong equation for that. The straightforward way to find the acceleration is to draw two separate free body diagrams (FBD) and get two separate equations, one for the two masses on the incline and one for the hanging mass. Once you have the common acceleration, you can find the tension between the masses by drawing a FBD for the 6.0-kg mass.
 
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  • #3
Unfortunately the image of the question is cropped at the right side. As a result, the answer to the question as posted is "a rope".
 
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FAQ: Atwood at an incline accelerating down

What is an Atwood machine, and how does it work on an incline?

An Atwood machine typically consists of two masses connected by a string that passes over a pulley. When placed on an incline, one or both masses are subjected to gravitational force components along the plane of the incline. This setup allows for the study of the interplay between gravitational force, tension in the string, and the incline's angle, resulting in the acceleration of the system.

How do you calculate the acceleration of the masses in an Atwood machine on an incline?

To calculate the acceleration, you need to consider the forces acting on each mass. For a mass m1 on an incline at angle θ, the forces include the gravitational component m1g sin(θ) along the incline and the tension T in the string. For a hanging mass m2, the forces are m2g and the tension T. Using Newton's second law, you can set up the equations: m1g sin(θ) - T = m1a and T - m2g = -m2a. Solving these simultaneous equations gives the acceleration a = (m2g - m1g sin(θ)) / (m1 + m2).

What role does the angle of the incline play in the system's acceleration?

The angle of the incline, θ, affects the component of gravitational force acting along the plane of the incline. As θ increases, the component m1g sin(θ) increases, which in turn affects the net force and therefore the acceleration of the system. A steeper incline results in a larger component of gravitational force along the incline, increasing the acceleration of the system.

How does the tension in the string change when the Atwood machine is on an incline?

The tension in the string is influenced by the forces acting on both masses and the angle of the incline. For a mass m1 on an incline, the tension T must balance the component of gravitational force along the incline and the acceleration of the mass. For a hanging mass m2, the tension must balance the gravitational force and the acceleration in the opposite direction. The tension can be found by solving the system of equations derived from Newton's second law for both masses.

What assumptions are made in analyzing an Atwood machine on an incline?

Several assumptions are typically made for simplicity: the pulley is frictionless and massless, the string is inextensible and massless, the incline is frictionless, and the gravitational field is uniform. These assumptions allow for straightforward application of Newton's laws to determine the forces and accelerations in the system.

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