Mechanics As Level Connected Particles

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prinqle
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#1
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#1
when working out tension, only one object will give the correct answer whilst the other one (despite taking to account all forces), never works. am i missing a crucial rule or something because ive about had it with connected particles and the concepts.
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ThomH97
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Can you give an example of a question where the approaches give different answers?
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prinqle
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(Original post by ThomH97)
Can you give an example of a question where the approaches give different answers?
perhaps im mentally slow but here. also an example in my textbook involves removing the objects and calculating force exerted using only tension and the scale pan- ive attached this as well (question c). i am beyond lost.
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Pangol
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(Original post by prinqle)
perhaps im mentally slow but here. also an example in my textbook involves removing the objects and calculating force exerted using only tension and the scale pan- ive attached this as well (question c). i am beyond lost.
In your working for the car, you are taking T and the resistive force as positive and the driving force as negative. You also take the acceleration as positive. But T and the resistive force point in the opposite direction to the acceleration, so they should be negative. Similarly, the driving force should be positive.

It often helps to draw a big arrow next to your diagram pointing in what you have decided is the positive direction, and then make sure you are following this convention.
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jamiecjx
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(Original post by prinqle)
perhaps im mentally slow but here. also an example in my textbook involves removing the objects and calculating force exerted using only tension and the scale pan- ive attached this as well (question c). i am beyond lost.
You've written down T+300-1500=500\times \frac{10}{7}
However you've taken Left as the positive direction on the left hand side, whereas on the Right hand side the direction you've taken as positive is Right
The actual equation should be 1500-300-T=500\times \frac{10}{7} if right is the positive direction
T is negative as it is pulling in the opposite direction to the direction of travel


while i was writing this pangol beat me to it
Last edited by jamiecjx; 2 years ago
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ThomH97
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For your handwritten example, you have acceleration the wrong way. Without seeing the question, I would imagine the whole system is accelerating to the right? If so, the 1500 driving force is in the same direction as the 10/7 acceleration. I'm also not sure how you've got from T+300-1500=500x10/7 (signs are wrong here anyway) to the RHS evaluating to 1914.3.

If you could post a whole question and your working through it, that would be better for identifying issues. All I can say at the moment is a (quite common) sign slip and a (rather drastic) numerical slip.
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