# Application of vectors (1 Viewer)

#### poptarts12345

##### Member
Hey, could someone explain to me what tension is and does it oppose motion.
For example in this question :
Is tension moving up the rope from the pink block or down the rope to the pink block?
Also why is the equation 3a=T-3gsin θ instead of 3a=3gsin θ -T

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#### DrDawn

##### Member
Also why is the equation 3a=T-3gsin θ instead of 3a=3gsin θ -T
That's what I find confusing with these types of questions also, can someone explain it please?

#### DrDawn

##### Member
Hey, could someone explain to me what tension is and does it oppose motion.
For example in this question :
Is tension moving up the rope from the pink block or down the rope to the pink block?
Also why is the equation 3a=T-3gsin θ instead of 3a=3gsin θ -T
View attachment 29594
Btw what textbook are you using?

#### idkkdi

##### Well-Known Member
Hey, could someone explain to me what tension is and does it oppose motion.
For example in this question :
Is tension moving up the rope from the pink block or down the rope to the pink block?
Also why is the equation 3a=T-3gsin θ instead of 3a=3gsin θ -T
View attachment 29594
Tension acts away from the pink object, always acts away from objects which are being pulled. You can't push with tension.

3a = T-3gsintheta is a sum of forces.
Net force is 3a. This is from 3g sintheta (its weight acting down the ramp) and T which is tension acting opposite to 3g sintheta.
Since its moving upwards with 3a, obviously weight force has the negative sign and T has the positive sign.
Btw what textbook are you using?
this looks like Cambridge mx1

#### poptarts12345

##### Member
Tension acts away from the pink object, always acts away from objects which are being pulled. You can't push with tension.

3a = T-3gsintheta is a sum of forces.
Net force is 3a. This is from 3g sintheta (its weight acting down the ramp) and T which is tension acting opposite to 3g sintheta.
Since its moving upwards with 3a, obviously weight force has the negative sign and T has the positive sign.
why is tension larger than 3gsinthetha

#### poptarts12345

##### Member
3a up the plane. 3a >0.
wait so if acceleration is up the plane then tension is greater than mg and when acceleration is down the plane tension is smaller than mg?

so if the question about a dropping ball attached to a string, the acceleration is moving downwards so tension would be smaller?

ugh why do they need to add phys to maths

#### Trebla

##### Administrator
Administrator
wait so if acceleration is up the plane then tension is greater than mg and when acceleration is down the plane tension is smaller than mg?

so if the question about a dropping ball attached to a string, the acceleration is moving downwards so tension would be smaller?

ugh why do they need to add phys to maths
An easier way to think of this is that if the directions are the same then they are positively related and if they are opposite then they are negatively related.

When you resolve the gravitational force into its components (parallel/perpendicular to the plane), the direction of the component that is parallel to the plane is going down the plane. In contrast, the direction of the tension in the string is going up the plane.

In the last sentence of the question, you are told that the direction of the acceleration vector a is going up the plane. Since T is going up the plane and the gravitational vector is going down the plane, the relationship is:
3a = T - 3gsin θ

If the question instead said that acceleration vector was going down the plane then the signs would be reversed. If question doesn't say what the direction of a was, you actually would not know which direction the net force is going because it depends on what the numerical values of T and θ are.

It is not clear what you mean by tension being greater than mg, because it is not greater than mg but rather a fraction of mg. Also, tension is a concept only when the string is fully taut (not loose) as it is caused by the "stretching" of the string to oppose the natural motion of the object. Tension is basically non-existent if you are dropping a ball vertically attached to a string, until it hits an equilibrium and the string is being stretched enough to completely oppose the natural motion of the ball.

#### fan96

##### 617 pages
wait so if acceleration is up the plane then tension is greater than mg and when acceleration is down the plane tension is smaller than mg?
The block accelerates in one dimension - either up or down the plane.
The numbers we use to describe acceleration are also one dimensional - either positive or negative.
It's simply a matter of choosing which plane direction is "positive" and which one is "negative".
The choice itself doesn't matter - you just need to be consistent.

The question says acceleration is $\bg_white a\, \text{m/s}$ up the plane, which is a suggestion that we should pick "up the plane" as positive.
In this case, tension is positive (pulling the block up the plane) and gravity is negative.
$\bg_white 3a = T-3g \sin \theta$.

You are also perfectly free to choose "down the plane" as positive acceleration. Then your equation becomes
$\bg_white 3a = 3g \sin \theta - T$.

This is essentially the same thing - the only difference is the sign of $\bg_white a$ will be flipped.
(You can make the substitution $\bg_white a \mapsto -a$ to see for yourself.)
So if you got $\bg_white a = 1$ with the first equation, you would get $\bg_white a = -1$ with the second.

This is of course what we expect - the block is either moving up or down the plane, so if we change our choice of which direction is "positive" then the acceleration will flip signs so that the block is still physically moving in the same direction.

That is, if $\bg_white a = 1$ with the first equation it means that the block is moving up the slope, as we have chosen a positive number to correspond to moving up the slope. If we instead choose a negative number to correspond to moving up the slope (which is the second equation), then we should expect to find $\bg_white a = -1$.

Note that the magnitude of $\bg_white T$ and $\bg_white 3g \sin \theta$ are always the same for this scenario. One becomes (greater than/smaller than) the other only because their signs flip when you change your choice of direction for positive acceleration.

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#### idkkdi

##### Well-Known Member
The block accelerates in one dimension - either up or down the plane.
The numbers we use to describe acceleration are also one dimensional - either positive or negative.
It's simply a matter of choosing which plane direction is "positive" and which one is "negative".
The choice itself doesn't matter - you just need to be consistent.

The question says acceleration is $\bg_white a\, \text{m/s}$ up the plane, which is a suggestion that we should pick "up the plane" as positive.
In this case, tension is positive (pulling the block up the plane) and gravity is negative.
$\bg_white 3a = T-3g \sin \theta$.

You are also perfectly free to choose "down the plane" as positive acceleration. Then your equation becomes
$\bg_white 3a = 3g \sin \theta - T$.

This is essentially the same thing - the only difference is the sign of $\bg_white a$ will be flipped.
(You can make the substitution $\bg_white a \mapsto -a$ to see for yourself.)
So if you got $\bg_white a = 1$ with the first equation, you would get $\bg_white a = -1$ with the second.

This is of course what we expect - the block is either moving up or down the plane, so if we change our choice of which direction is "positive" then the acceleration will flip signs so that the block is still physically moving in the same direction.

That is, if $\bg_white a = 1$ with the first equation it means that the block is moving up the slope, as we have chosen a positive number to correspond to moving up the slope. If we instead choose a negative number to correspond to moving up the slope (which is the second equation), then we should expect to find $\bg_white a = -1$.

Note that the magnitude of $\bg_white T$ and $\bg_white 3g \sin \theta$ are always the same for this scenario. One becomes (greater than/smaller than) the other only because their signs flip when you change your choice of direction for positive acceleration.
normally we can choose the positive direction, but here its clearly stated 3a is up, so @poptarts12345 use the given direction for this q.

#### fan96

##### 617 pages
normally we can choose the positive direction, but here its clearly stated 3a is up, so @poptarts12345 use the given direction for this q.
The question only says the object is accelerating $\bg_white a$ m/s up the plane.
There is no positive direction defined here at all.

Regardless, it doesn't matter which direction is chosen to be positive. The only thing that matters is that you clearly state which convention you have chosen and that you are consistent with your choice.

#### fan96

##### 617 pages
One more thing:
Hey, could someone explain to me what tension is and does it oppose motion.
Tension is a property of taut strings/ropes etc. that pull attached objects towards each other.

Suppose you attach a ball to a string and then hold the string up midair, suspending the ball.

The ball is not moving, despite the pull of gravity. That's because there's a force provided by the string (via your hand) pulling it up. We call this tension.

At the same time, you're also having to do slightly more work than usual in trying to keep your hand in the air. The reason is that the string (with ball attached) is pulling down equally as much on your hand. That's also tension.

In the real world it's not quite this simple because real-life strings have mass, which complicates things. But in simple problems like these we usually assume strings and ropes have no mass.