Newton's Second Law in NON-inertial frame of reference

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To determine the acceleration of a frame where the tension in cord A is twice that in cord B, the forces acting on the steel ball must be analyzed in the context of a non-inertial reference frame. The angles of the cords are both 60 degrees, and the tension in the cords can be expressed in terms of the weight of the ball. By breaking down the components of the tensions and applying Newton's second law, the relationship between the tensions and the acceleration of the frame can be established. The problem requires redrawing the force diagram to account for the acceleration, allowing for the correct calculation of the angle and the resulting tension relationship. Understanding these dynamics is crucial for solving the problem accurately.
Spiewgels
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Homework Statement


The steel ball is suspended from the accelerating frame by the two cords A and B. The angles (they are on the inside) are both 60 degrees.

Determine the acceleration of the frame which will cause the tension in A to be twice that in B. The acceleration is going to the right and the cord A is to the right of the moving frame.

Provide your answer in m/s/s with one decimal point accuracy


Homework Equations


I want to know how to relate the forces on the inside of the accelerating frame to the accelerating frame itself.


The Attempt at a Solution


Thus far, I have drawn free body diagrams to the inside cords and steel ball. I broke down the components of cord A and B and found the x-coordinate of cord A to be Acos60 and y-coordinate of Asin60. I got these same results for cord B. Combining knowns I've determined both cords tension to be .87w where w equals the weight of the ball. I'm now stuck and don't know how this relates to the moving frame where I think the force is F=ma(of x) and a(of x)=F/m...Have I screwed up this entire problem?
 
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You've apparently solved the problem of a ball hanging in a gravitational field using F=mg. In an frame with acceleration vector a, the equivalent force is given by F=ma (obviously) in the direction opposite to the acceleration. So redraw your force diagram, but this time instead of drawing the 'external' force as pointing straight down, let it point at some angle. Your job is to determine that angle so you get the right tension relation.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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