Two boxes connected by strings at an angle

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Two boxes, A and B, connected by a massless string at a 7° angle, require a force to achieve an acceleration of 0.2 m/s². Box A has a mass of 9 kg and box B has a mass of 4 kg, leading to a total mass of 13 kg. The tension force in box A is calculated using Ftx = Ftcos23°, and the required force was initially miscalculated but corrected to 3.04 N. The angle of the string between the boxes does not affect the acceleration, as it is considered an internal force. To find the tension in the external string, the 7° angle must be included in the calculations.
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Homework Statement


Two boxes, A & B are on a flat surface, and are connected by a massless string at an angle of 7° to the horizontal. Box A masses 9kg, box b 4kg. The boxes are pulled by a string connected to box A at an angle of 23° to the horizontal. Find the force required in order to achieve an acceleration of .2m/s^2. Assume no frictional force.
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Homework Equations


Fnet = m(a)


The Attempt at a Solution


So far I've figured out the tension force on box A in the horizontal direction to be Ftx = Ftcos23°.
Normally, I would just multiply m(a), in this case 13(.2) to find 2.8N required in the horizontal direction. Given that Ftx = Ftcos23°, I would think to divide by cos23°, but this results in a negative answer. I'm also unsure as to how the boxes being connected by an angled string affects this.

EDIT: The negative answer was a calculation error, I'm getting 3.04N now.
 
Last edited:
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Yes, except 13(.2) = 2.6 N.:wink:

The nice thing about free body diagrams of the system of 2 boxes and the string between them is that only forces external to the system are considered. The acceleration is independent of the angle of the string in between the boxes, because the string between the boxes is an internal part of the system. if you needed to find the tension in that external string, then you must include the 7 degree angle in your solution for that tensile force.
 
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