Hi, please could i have help on this question?

I’m not sure why the markscheme uses 2.4 as the resultant force in f=ma? Surely if there is 2.4N force the resultant force is the line i drew in red and we use that resultant force in f=ma?

Here is the question: https://app.gemoo.com/share/image-annotation/638917612761178112?codeId=Pal134WXA2p93&origin=imageurlgenerator

Thank you!!

I’m not sure why the markscheme uses 2.4 as the resultant force in f=ma? Surely if there is 2.4N force the resultant force is the line i drew in red and we use that resultant force in f=ma?

Here is the question: https://app.gemoo.com/share/image-annotation/638917612761178112?codeId=Pal134WXA2p93&origin=imageurlgenerator

Thank you!!

2.5 N force is acting at angle 'x' (theta).If you resolve the force vertically then F= 2.5 cos x and horizontallly 2.4 = 2.5 sin x.

This gives us sin x = 24/25

If 2.4 force is removed then only force left horizontally is horizontal component of 2.5 N which is 2.5 sin x = 2.5 × 24/25 = 2.4 N

Since movement is in horizontal direction this is equal to m.a

Therefore m.a = 2.4 N

This gives us sin x = 24/25

If 2.4 force is removed then only force left horizontally is horizontal component of 2.5 N which is 2.5 sin x = 2.5 × 24/25 = 2.4 N

Since movement is in horizontal direction this is equal to m.a

Therefore m.a = 2.4 N

Original post by Justcurious_yr12

2.5 N force is acting at angle 'x' (theta).If you resolve the force vertically then F= 2.5 cos x and horizontallly 2.4 = 2.5 sin x.

This gives us sin x = 24/25

If 2.4 force is removed then only force left horizontally is horizontal component of 2.5 N which is 2.5 sin x = 2.5 × 24/25 = 2.4 N

Since movement is in horizontal direction this is equal to m.a

Therefore m.a = 2.4 N

This gives us sin x = 24/25

If 2.4 force is removed then only force left horizontally is horizontal component of 2.5 N which is 2.5 sin x = 2.5 × 24/25 = 2.4 N

Since movement is in horizontal direction this is equal to m.a

Therefore m.a = 2.4 N

Thank you! So just to check, the vertical component of 2.5 is balanced by F so when the 2.4 force is removed, the resultant force is the 2.4N which is the horizontal component of F and that is the resultant force?

I was also trying to attempt this question: https://app.gemoo.com/share/image-annotation/639108670275133440?codeId=v67gqQNnpd3Jl&origin=imageurlgenerator

But i was unsure how to resolve it? We don’t know the angle between 6N and the ground or 10N and the ground so i didn’t know how to split it into the horizontal and vertical components? I was even more confused when the marscheme used the angle 70 to resolve, I didn’t know where they got that from? Please could i have some help, thank you so much!!!

But i was unsure how to resolve it? We don’t know the angle between 6N and the ground or 10N and the ground so i didn’t know how to split it into the horizontal and vertical components? I was even more confused when the marscheme used the angle 70 to resolve, I didn’t know where they got that from? Please could i have some help, thank you so much!!!

Original post by anonymous294

Thank you! So just to check, the vertical component of 2.5 is balanced by F so when the 2.4 force is removed, the resultant force is the 2.4N which is the horizontal component of F and that is the resultant force?

The force 'F' is a force acting at 90 degrees to horizontal and therefore will not have any horizontal component.

2.5 N force is acting at an angle theta where sin (theta)= 25/24. Since it is acting at an angle it will have vertical and horizontal component.

As the movement is in horizontal direction only the horizontal component of 2.5 N will be used.'F' being perpendicular to the direction of movement is ignored.

Original post by anonymous294

I was also trying to attempt this question: https://app.gemoo.com/share/image-annotation/639108670275133440?codeId=v67gqQNnpd3Jl&origin=imageurlgenerator

But i was unsure how to resolve it? We don’t know the angle between 6N and the ground or 10N and the ground so i didn’t know how to split it into the horizontal and vertical components? I was even more confused when the marscheme used the angle 70 to resolve, I didn’t know where they got that from? Please could i have some help, thank you so much!!!

But i was unsure how to resolve it? We don’t know the angle between 6N and the ground or 10N and the ground so i didn’t know how to split it into the horizontal and vertical components? I was even more confused when the marscheme used the angle 70 to resolve, I didn’t know where they got that from? Please could i have some help, thank you so much!!!

C²= a²+b² - 2 ab Cos c where Cos C is Cos 110

and a= 10 and b=6

Original post by Justcurious_yr12

You can use the cosine formula to solve it

C²= a²+b² - 2 ab Cos c where Cos C is Cos 110

and a= 10 and b=6

C²= a²+b² - 2 ab Cos c where Cos C is Cos 110

and a= 10 and b=6

I used that but they said that the angle isn’t 110, it’s 70?

Original post by anonymous294

I used that but they said that the angle isn’t 110, it’s 70?

My bad: it is 70 degree actually. I did not pay attention to direction of vector. If you move the 6N vector so that it's tail touches the tip of 10 N vector then angle becomes 180-110= 70 degrees.

Use

C²= a²+b²-2ab cos 70

If you do not want to move the 6N vector then you have to consider it's magnitude as negative

C²= a²+b²- 2 a (-b) cos 110

Where B = 6N and a =10 N

Original post by Justcurious_yr12

My bad: it is 70 degree actually. I did not pay attention to direction of vector. If you move the 6N vector so that it's tail touches the tip of 10 N vector then angle becomes 180-110= 70 degrees.

Use

C²= a²+b²-2ab cos 70

If you do not want to move the 6N vector then you have to consider it's magnitude as negative

C²= a²+b²- 2 a (-b) cos 110

Where B = 6N and a =10 N

Use

C²= a²+b²-2ab cos 70

If you do not want to move the 6N vector then you have to consider it's magnitude as negative

C²= a²+b²- 2 a (-b) cos 110

Where B = 6N and a =10 N

Thank you, but im quite confused on how this changes? As in normally we have to resolve vertically and horizontally to find the resultant force but here we are joining the two vectors up?

Original post by anonymous294

Thank you, but im quite confused on how this changes? As in normally we have to resolve vertically and horizontally to find the resultant force but here we are joining the two vectors up?

Resolving vertically and horizontally is the preferred method and you should always do that if you have the angle between the vector and the axis/plane. In this particular question you have an angle between 2 vectors. You can still resolve it vertically and horizontally but the method is long and complex. These kind of questions are rare and would mostly ask for resultant force between the two. If in a question you have to find acceleration/ tension/ velocity then it would be a straightforward horizontal and vertical component.

Original post by Justcurious_yr12

Resolving vertically and horizontally is the preferred method and you should always do that if you have the angle between the vector and the axis/plane. In this particular question you have an angle between 2 vectors. You can still resolve it vertically and horizontally but the method is long and complex. These kind of questions are rare and would mostly ask for resultant force between the two. If in a question you have to find acceleration/ tension/ velocity then it would be a straightforward horizontal and vertical component.

Ok thank you so much!!

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