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so, to create an equation. OP is the positive direction?

the force 8x is acting in the opposite direction - to PO, so I guess it's travelling back to O?

also, there's a resistive force - I assume thats resistant to the positive direction of OP - therefore acting in the direction of PO?

how do I model this equation - I assume I start with f=ma?

however, f can't be negative?

the force 8x is acting in the opposite direction - to PO, so I guess it's travelling back to O?

also, there's a resistive force - I assume thats resistant to the positive direction of OP - therefore acting in the direction of PO?

how do I model this equation - I assume I start with f=ma?

however, f can't be negative?

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so, to create an equation. OP is the positive direction?

the force 8x is acting in the opposite direction - to PO, so I guess it's travelling back to O?

also, there's a resistive force - I assume thats resistant to the positive direction of OP - therefore acting in the direction of PO?

how do I model this equation - I assume I start with f=ma?

however, f can't be negative?

**FurtherMaths2020**)so, to create an equation. OP is the positive direction?

the force 8x is acting in the opposite direction - to PO, so I guess it's travelling back to O?

also, there's a resistive force - I assume thats resistant to the positive direction of OP - therefore acting in the direction of PO?

how do I model this equation - I assume I start with f=ma?

however, f can't be negative?

Can you create an F = ma equation using all this? Even if you're unsure please post your attempt and we can help you if it's wrong.

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You are told that initially the particle is "moving in the direction of increasing x" so if you draw a force diagram, the motion will be to the right. You are correct that the 8x force acts in a direction back to O so that will be pointing to the left. A resistance force always acts in the opposite direction to the motion so it must be equal to -4v (this force will change direction depending on the direction that the particle is moving in).

Can you create an F = ma equation using all this? Even if you're unsure please post your attempt and we can help you if it's wrong.

**Sir Cumference**)You are told that initially the particle is "moving in the direction of increasing x" so if you draw a force diagram, the motion will be to the right. You are correct that the 8x force acts in a direction back to O so that will be pointing to the left. A resistance force always acts in the opposite direction to the motion so it must be equal to -4v (this force will change direction depending on the direction that the particle is moving in).

Can you create an F = ma equation using all this? Even if you're unsure please post your attempt and we can help you if it's wrong.

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but the force in my working out is acting in the wrong direction - so it's incorrect but you can't have a negative force, so how do I resolve this?

**FurtherMaths2020**)but the force in my working out is acting in the wrong direction - so it's incorrect but you can't have a negative force, so how do I resolve this?

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You can have a negative force since a force is a vector quantity - it depends which direction you take as positive. The x that we're interested in is positive in the direction OP so we should take right as positive. Since the two forces are acting in an opposite direction to x increasing, the resultant force is equal to (-8x) + (-4v) which means that the resultant force is negative. This makes sense because the resultant force is pointing to the left initially, in the opposite direction to motion. Can you fix your equation?

**Sir Cumference**)You can have a negative force since a force is a vector quantity - it depends which direction you take as positive. The x that we're interested in is positive in the direction OP so we should take right as positive. Since the two forces are acting in an opposite direction to x increasing, the resultant force is equal to (-8x) + (-4v) which means that the resultant force is negative. This makes sense because the resultant force is pointing to the left initially, in the opposite direction to motion. Can you fix your equation?

yes, I have taken right to be positive.

yes, the forces are negative but I was told that force can never be negative. as in, it can't be: -f = ma

the answer has in the textbook is -f = ma, or is it we can never have negative acceleration in the case of f = ma?

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so, in all f = ma equations, I can have a negative force?

yes, I have taken right to be positive.

yes, the forces are negative but I was told that force can never be negative. as in, it can't be: -f = ma

the answer has in the textbook is -f = ma, or is it we can never have negative acceleration in the case of f = ma?

**FurtherMaths2020**)so, in all f = ma equations, I can have a negative force?

yes, I have taken right to be positive.

yes, the forces are negative but I was told that force can never be negative. as in, it can't be: -f = ma

the answer has in the textbook is -f = ma, or is it we can never have negative acceleration in the case of f = ma?

-F = ma would suggest that the resultant force is in an opposite direction to the acceleration which is impossible.

__The equation is always F = ma.__

But F and a within the equation can be negative.

In this case take right as positive and work out the resultant force which is equal to

(-8x) + (-4v)

and then using the equation f = ma you have

-8x - 4v = ma

Does this make sense?

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(Original post by

Yes if the resultant force is in an opposite direction to your positive direction then the resultant force and the acceleration are negative. Force/acceleration can be positive or negative.

-F = ma would suggest that the resultant force is in an opposite direction to the acceleration which is impossible.

But F and a within the equation can be negative.

In this case take right as positive and work out the resultant force which is equal to

(-8x) + (-4v)

and then using the equation f = ma you have

-8x - 4v = ma

Does this make sense?

**Sir Cumference**)Yes if the resultant force is in an opposite direction to your positive direction then the resultant force and the acceleration are negative. Force/acceleration can be positive or negative.

-F = ma would suggest that the resultant force is in an opposite direction to the acceleration which is impossible.

__The equation is always F = ma.__But F and a within the equation can be negative.

In this case take right as positive and work out the resultant force which is equal to

(-8x) + (-4v)

and then using the equation f = ma you have

-8x - 4v = ma

Does this make sense?

what would be the resultant force (the positive force acting in the same direction as the acceleration - to the right) in this case?

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#8

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ok, so the resultant force and acceleration are acting in the positive direction (to the right), but the 2 forces we're using (which aren't the resultant force) are acting in the opposite direction and therefore are individually negative?

**FurtherMaths2020**)ok, so the resultant force and acceleration are acting in the positive direction (to the right), but the 2 forces we're using (which aren't the resultant force) are acting in the opposite direction and therefore are individually negative?

-8x - 4v = ma

Firstly, does this equation make sense now? If it doesn't make sure you say. x and v are positive quantities (in the direction of x increasing) which means that -8x and -4v are negative quantities which means that ma (equal to the resultant force) is negative and therefore the acceleration is negative. But even without considering this equation, it should be clear that the all the forces are acting in a negative direction and therefore so is the resultant force & acceleration.

Last edited by Sir Cumference; 1 month ago

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#9

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ok, so the resultant force and acceleration are acting in the positive direction (to the right), but the 2 forces we're using (which aren't the resultant force) are acting in the opposite direction and therefore are individually negative?

what would be the resultant force (the positive force acting in the same direction as the acceleration - to the right) in this case?

**FurtherMaths2020**)ok, so the resultant force and acceleration are acting in the positive direction (to the right), but the 2 forces we're using (which aren't the resultant force) are acting in the opposite direction and therefore are individually negative?

what would be the resultant force (the positive force acting in the same direction as the acceleration - to the right) in this case?

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(Original post by

No, the resultant force must be acting in the negative direction because the two forces are acting in the negative direction. The acceleration is also acting in a negative direction. Look at the equation again:

-8x - 4v = ma

Firstly, does this equation make sense now? If it doesn't make sure you say. x and v are positive quantities (in the direction of x increasing) which means that -8x and -4v are negative quantities which means that ma (equal to the resultant force) is negative and therefore the acceleration is negative. But even without considering this equation, it should be clear that the all the forces are acting in a negative direction and therefore so is the resultant force & acceleration.

**Sir Cumference**)No, the resultant force must be acting in the negative direction because the two forces are acting in the negative direction. The acceleration is also acting in a negative direction. Look at the equation again:

-8x - 4v = ma

Firstly, does this equation make sense now? If it doesn't make sure you say. x and v are positive quantities (in the direction of x increasing) which means that -8x and -4v are negative quantities which means that ma (equal to the resultant force) is negative and therefore the acceleration is negative. But even without considering this equation, it should be clear that the all the forces are acting in a negative direction and therefore so is the resultant force & acceleration.

so, if when f = ma, and the force is positive on the LHS, but negative if moved to the RHS, that would be when the particle is traveling in the positive direction, but the acceleration is still negative?

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(Original post by

I should add that in SHM eventually the particle will be travelling in the opposite direction (back to O). When this happens, v will be negative but the

**Sir Cumference**)I should add that in SHM eventually the particle will be travelling in the opposite direction (back to O). When this happens, v will be negative but the

**resistive force will still be (-4v) because it is always in an opposite direction to motion**. Also x will still be positive so the other force will still be (-8x) which means that the equation -8x - 4v = ma is still satisfied.ah, ok - a resistive force in SHM is always resisting direction of motion?

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