Hi, so I've been doing resultant forces using vectors and I'm confused on something.

I get the whole vertical Fsinx⁰ and horizontal Fcosx⁰, but in some of the practice questions (AQA) they're using Fcosx⁰ for one of the forces for vertical resolutions, and vice versa.

I've looked through the book countless times and can't find anything explaining any rule on why we sometimes need to use the opposite function when resolving forces.

Any help appreciated. Thankyou.

I get the whole vertical Fsinx⁰ and horizontal Fcosx⁰, but in some of the practice questions (AQA) they're using Fcosx⁰ for one of the forces for vertical resolutions, and vice versa.

I've looked through the book countless times and can't find anything explaining any rule on why we sometimes need to use the opposite function when resolving forces.

Any help appreciated. Thankyou.

(edited 1 month ago)

Original post by Rgl7420

Hi, so I've been doing resultant forces using vectors and I'm confused on something.

I get the whole vertical Fsinx⁰ and horizontal Fcosx⁰, but in some of the practice questions (AQA) they're using Fcosx⁰ for one of the forces for vertical resolutions, and vice versa.

I've looked through the book countless times and can't find anything explaining any rule on why we sometimes need to use the opposite function when resolving forces.

Any help appreciated. Thankyou.

I get the whole vertical Fsinx⁰ and horizontal Fcosx⁰, but in some of the practice questions (AQA) they're using Fcosx⁰ for one of the forces for vertical resolutions, and vice versa.

I've looked through the book countless times and can't find anything explaining any rule on why we sometimes need to use the opposite function when resolving forces.

Any help appreciated. Thankyou.

you probably need to post the question you're working on for specific help, but note that sin x = cos(90 - x), so make sure that they're referring to the same angle you're thinking of

Hi, thanks for the reply.

I should have included an example in the first place, but I've edited my post to show an example. I get the logic of how you solve for F and everything, but it's resolving in the y-direction, but only the 12N 18⁰ force uses cos instead of sin, the rest use sin (as the book tells me to do when resolving vertically). If you could take a look at that image I'd really appreciate it.

I should have included an example in the first place, but I've edited my post to show an example. I get the logic of how you solve for F and everything, but it's resolving in the y-direction, but only the 12N 18⁰ force uses cos instead of sin, the rest use sin (as the book tells me to do when resolving vertically). If you could take a look at that image I'd really appreciate it.

Original post by Rgl7420

Hi, thanks for the reply.

I should have included an example in the first place, but I've edited my post to show an example. I get the logic of how you solve for F and everything, but it's resolving in the y-direction, but only the 12N 18⁰ force uses cos instead of sin, the rest use sin (as the book tells me to do when resolving vertically). If you could take a look at that image I'd really appreciate it.

I should have included an example in the first place, but I've edited my post to show an example. I get the logic of how you solve for F and everything, but it's resolving in the y-direction, but only the 12N 18⁰ force uses cos instead of sin, the rest use sin (as the book tells me to do when resolving vertically). If you could take a look at that image I'd really appreciate it.

That looks perfectly natural to me - they've given 18 as the angle between the force vector and the y-axis so it's natural to use the cos of that angle,

It might be easier to think about what you would do if you were resolving along the x-axis - all the other angles given are between the force vectors and the x-axis, so you would be using cos 20, cos 30 and cos 60. Because the question asks for resolution with the y-axis, you're actually using sin 20, sin30 and sin 60

Original post by davros

That looks perfectly natural to me - they've given 18 as the angle between the force vector and the y-axis so it's natural to use the cos of that angle,

It might be easier to think about what you would do if you were resolving along the x-axis - all the other angles given are between the force vectors and the x-axis, so you would be using cos 20, cos 30 and cos 60. Because the question asks for resolution with the y-axis, you're actually using sin 20, sin30 and sin 60

It might be easier to think about what you would do if you were resolving along the x-axis - all the other angles given are between the force vectors and the x-axis, so you would be using cos 20, cos 30 and cos 60. Because the question asks for resolution with the y-axis, you're actually using sin 20, sin30 and sin 60

Okay, I think I get it. Because 12cos18⁰ is the same as 12sin72⁰. So it's what axis the angle comes from that determines what function to use?

Is that due to how the vector operates? Like with that 12N 18⁰ force, the y-axis essentially acts as the horizontal component, while the vertical component is horizontal relative to the triangle produced from the component vectors?

Original post by Rgl7420

Okay, I think I get it. Because 12cos18⁰ is the same as 12sin72⁰. So it's what axis the angle comes from that determines what function to use?

Is that due to how the vector operates? Like with that 12N 18⁰ force, the y-axis essentially acts as the horizontal component, while the vertical component is horizontal relative to the triangle produced from the component vectors?

Is that due to how the vector operates? Like with that 12N 18⁰ force, the y-axis essentially acts as the horizontal component, while the vertical component is horizontal relative to the triangle produced from the component vectors?

Pretty much - you're basically doing what's called a "projection" of one vector onto another to get a component in that direction. If they'd labelled the 72 degree angle on the diagram you'd have known exactly what to do without thinking, and since cos 72 = sin 18 and sin 72 = cos 18, the results come out the same!

(edited 1 month ago)

Original post by davros

Pretty much - you're basically doing what's called a "projection" of one vector onto another to get a component in that direction. If they'd labelled the 72 degree angle on the diagram you'd have known exactly what to do without thinking, and since cos 72 = sin 18 and sin 72 = cos 18, the results come out the same!

Thankyou very much, you're an absolute lifesaver.

Hey, i see why you may get stuck practicing this because resolving forces is a topic that you get used to after practicing many questions on it. Trust me I know. For this question, let’s consider the bottom force of 12N. All forces have a horizontal and vertical component so the 12N can be resolved (resolved means split into its horizontal and vertical components). Using trigonometry, the horizontal component of the 12N is 12Sin(18) which acts to the left because the 12N is pointing left. Whereas the vertical component of the 12N force is 12Cos(18). If you want to see this visually draw the triangle and label sides hypotenuse opposite and adjacent to see trig that I’m doing. The reason some questions use the Cos(x) is because it depends on where the force is pointing. Like for the 12N force the vertical component uses cos but for the 6N force the vertical component is 6sin(30) because the opposite side = (6N) x sin(30). It’s just about the direction the force points and using trig to find the force that acts in that direction. I hope this helps.

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