This is the question: https://www.imghippo.com/i/L0pK81719761374.png

I wasn't sure on part b) of this question and so I checked the mark scheme which made me even more confused. Could someone explain how you would go about answering part b) of this question?

I wasn't sure on part b) of this question and so I checked the mark scheme which made me even more confused. Could someone explain how you would go about answering part b) of this question?

Original post by TwisterBlade596

This is the question: https://www.imghippo.com/i/L0pK81719761374.png

I wasn't sure on part b) of this question and so I checked the mark scheme which made me even more confused. Could someone explain how you would go about answering part b) of this question?

I wasn't sure on part b) of this question and so I checked the mark scheme which made me even more confused. Could someone explain how you would go about answering part b) of this question?

Not sure what the mark scheme says, but the string is in tension and there will be two T's, one heading straight down and the other along the ramp. So add those two forces to get the resultant force on P

That was what I thought. The tension I calculated was 32.88N, which was on either side of P so I multiplied it by 2 to give 65.76 N.

This is what the model answer does: https://www.imghippo.com/i/RWhaI1719762163.png

This is what the model answer does: https://www.imghippo.com/i/RWhaI1719762163.png

Original post by TwisterBlade596

That was what I thought. The tension I calculated was 32.88N, which was on either side of P so I multiplied it by 2 to give 65.76 N.

This is what the model answer does: https://www.imghippo.com/i/RWhaI1719762163.png

This is what the model answer does: https://www.imghippo.com/i/RWhaI1719762163.png

If you have two forces of magnitude F, their sum will not have magnitude 2F unless they act in the same direction. You need to take account of the different directions (either use the cosine rule, or resolve horizontally and vertically).

Original post by DFranklin

If you have two forces of magnitude F, their sum will not have magnitude 2F unless they act in the same direction. You need to take account of the different directions (either use the cosine rule, or resolve horizontally and vertically).

Ok. I think I understand the cosine part but I then tried resolving horizontally and vertically but I did not get the same answer compared to when using cosine.

Original post by TwisterBlade596

Ok. I think I understand the cosine part but I then tried resolving horizontally and vertically but I did not get the same answer compared to when using cosine.

if you resolve horizontally and vertically and add to get the net horizontal and vertical components, then use pythagoras you should get the same ans.

(edited 1 month ago)

Original post by mqb2766

if you resolve horizontally and vertically and add to get the net horizontal and vertical components, then use pythagoras you should get the same ans.

I can only find one vertical and one horizontal component (I'm assuming there may be more because you have mentioned addition?), which may explain why I am still getting a different answer.

(edited 1 month ago)

Original post by TwisterBlade596

I can only find one vertical and one horizontal component (I'm assuming there may be more because you have mentioned addition?), which may explain why I am still getting a different answer.

Youd have to upload what you did. The downwards T would just be (0,T) and the T down ramp would be (4T/5, 3T/5). Then add the components and use pythagoras, thougth using the 2Tcos(theta) isosceles triangle is obviously the simplest thing to do.

(edited 1 month ago)

Original post by mqb2766

Youd have to upload what you did. The downwards T would just be (0,T) and the T down ramp would be (4T/5, 3T/5). Then add the components and use pythagoras, thougth using the 2Tcos(theta) isosceles triangle is obviously the simplest thing to do.

https://www.imghippo.com/i/kNOFb1719770169.jpg

Original post by TwisterBlade596

You "cant" have a horizontal component which is larger than the force.

The force is always the hypotenuse and the horizontal and vertical componets are the legs of the right triangle. So the tension down the ramp is resolved as (4T/5, 3T/5) as your triangle is a 3:4:5 and the tension downwards is (resolved as) (0,T). Then add each vector together and get the magnitude using pythagoras.

(edited 1 month ago)

Original post by mqb2766

You "cant" have a horizontal component which is larger than the force.

The force is always the hypotenuse and the horizontal and vertical componets are the legs of the right triangle. So the tension down the ramp is resolved as (4T/5, 3T/5) as your triangle is a 3:4:5 and the tension downwards is (resolved as) (0,T). Then add each vector together and get the magnitude using pythagoras.

The force is always the hypotenuse and the horizontal and vertical componets are the legs of the right triangle. So the tension down the ramp is resolved as (4T/5, 3T/5) as your triangle is a 3:4:5 and the tension downwards is (resolved as) (0,T). Then add each vector together and get the magnitude using pythagoras.

My horizontal component wasn't greater than the force...(I had 54.8 N as the force, which is greater than 32.88 )

I redrew the triangle with dimensions T, 4T/5 and 3T/5, but I don't entirely understand what you mean. What is the difference between the "tension down the ramp" and the "tension downwards"?

Original post by TwisterBlade596

My horizontal component wasn't greater than the force...(I had 54.8 N as the force, which is greater than 32.88 )

I redrew the triangle with dimensions T, 4T/5 and 3T/5, but I don't entirely understand what you mean. What is the difference between the "tension down the ramp" and the "tension downwards"?

I redrew the triangle with dimensions T, 4T/5 and 3T/5, but I don't entirely understand what you mean. What is the difference between the "tension down the ramp" and the "tension downwards"?

"downwards" is the tension in the vertical part of the string which just has a value T pointing vertically downwards

Tension "down the ramp" is at an angle alpha to the horizontal, so it has a component Tcos(alpha) horizontally and Tsin(alpha) vertically downwards.

The total horizontal component is clearly just Tcos(alpha). The net vertical component of the total force is the sum of the 2 vertical components given above.

So now just use Pythagoras to calculate the total force and you should get the same answer as in the mark scheme.

Original post by TwisterBlade596

My horizontal component wasn't greater than the force...(I had 54.8 N as the force, which is greater than 32.88 )

I redrew the triangle with dimensions T, 4T/5 and 3T/5, but I don't entirely understand what you mean. What is the difference between the "tension down the ramp" and the "tension downwards"?

I redrew the triangle with dimensions T, 4T/5 and 3T/5, but I don't entirely understand what you mean. What is the difference between the "tension down the ramp" and the "tension downwards"?

Im not sure what you were doing with 32/tan(alpha), then but ...

You have two forces, both of magnitude T. Theyre in different direction and you want to add them (adding vectors). If you resolve each tension (one is down towards B, one is down the ramp towards A), in the horizontal and vertical directions, then sum those as normal to get the resolved net force, then pythagoras to get the magnitude of the net force.

Resolving T (downwards towards B) is triviallly (0,T)

Resolving T (down the ramp towards A) is (4T/5, 3T/5)

so add those to get the resolved net force and do pythagoras. As above, its easier to do 2Tcos(theta)

(edited 1 month ago)

I think I slightly get it now...Could either of you draw a diagram so that I can properly visualise what is going on (because obviously my previous diagram was inaccurate)?

Original post by TwisterBlade596

I think I slightly get it now...Could either of you draw a diagram so that I can properly visualise what is going on (because obviously my previous diagram was inaccurate)?

Youre really just adding the two tensions tip to tail as usual and either using the fact that you get an isosceles triangle (model solution) or you consider the horizontal and vertical components of each tension and add those.

A quick diagram is something like

The red arrows are the two tensions the dotted one is the tip to tail addition, the black dashed are resolving the dotted one horizontally and vertically and the green dashed-dot is the resultant force (the base of the isosceles triangle of length 2Tcos(theta))

(edited 1 month ago)

Original post by mqb2766

Youre really just adding the two tensions tip to tail as usual and either using the fact that you get an isosceles triangle (model solution) or you consider the horizontal and vertical components of each tension and add those.

A quick diagram is something like

The red arrows are the two tensions the dotted one is the tip to tail addition, the black dashed are resolving the dotted one horizontally and vertically and the green dashed-dot is the resultant force (the base of the isosceles triangle of length 2Tcos(theta))

A quick diagram is something like

The red arrows are the two tensions the dotted one is the tip to tail addition, the black dashed are resolving the dotted one horizontally and vertically and the green dashed-dot is the resultant force (the base of the isosceles triangle of length 2Tcos(theta))

What is tip to tail addition?

Original post by TwisterBlade596

What is tip to tail addition?

When you add vectors. Its just another way of saying the two vectors define a parallelogram and the resultant (add them) is the diagonal.

https://www.varsitytutors.com/hotmath/hotmath_help/topics/adding-and-subtracting-vectors

You should have covered this, even the basics at gcse.

Oh ok I did not recognise that phrase. I've looked at the diagram and I think I do understand the resolving of the tensions. But I don't understand why the base of is 2Tcosalpha and not 2T^2 - 2Tcos(alpha+90) then square rooted.

(edited 1 month ago)

Original post by TwisterBlade596

Oh ok I did not recognise that phrase. I've looked at the diagram and I think I do understand the resolving of the tensions. But I don't understand why the base of is 2Tcosalpha and not 2T^2 - 2Tcos(alpha+90) then square rooted.

Its an isosceles triangle, so the perpendicular bisector makes 2 right triangles, where each base is Tcos(theta) so twice that. You can use the cos rule, but its a bit more work. Your second term should be T^2, not T, so when you root it you can take the T out and a bit of trig simplification would get you back to 2Tcos(theta)

(edited 1 month ago)

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