Physics AS question (resultant forces/vectors)

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KK.Violinist
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In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle 20º left of the forwards direction with force of 2250N. The second acts 15º to the right of direction with force of 2000N. Find resultant force on the stuck car.

I understand it but not sure on how to solve it. Do I apply the parallelogram rule and add the two forces?

Thanks in advance.
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hannaheliza27
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Could you do 2250 x cos20 + 2000 x cos15 ?
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joostan
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(Original post by KK.Violinist)
In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle 20º left of the forwards direction with force of 2250N. The second acts 15º to the right of direction with force of 2000N. Find resultant force on the stuck car.

I understand it but not sure on how to solve it. Do I apply the parallelogram rule and add the two forces?

Thanks in advance.
I'd start with a diagram modelling the car as a point, then break it down into horizontal and vertical components.
Then use |F|=\sqrt{F_{horizontal}^2+F_{vertical}^2}
And \tan(\theta)=\dfrac{F_{horizontal}}{F_{vertical}} for appropriate \theta
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KK.Violinist
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(Original post by hannaheliza27)
Could you do 2250 x cos20 + 2000 x cos15 ?
so that would be 4046.16N as resultant force. I'm not sure yet so I'll use the method given from the other post and get back to you
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KK.Violinist
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(Original post by joostan)
I'd start with a diagram modelling the car as a point, then break it down into horizontal and vertical components.
Then use |F|=\sqrt{F_{horizontal}^2+F_{vertical}^2}
And \tan(\theta)=\dfrac{F_{horizontal}}{F_{vertical}} for appropriate \theta
horizontal=
2000 x cos15 = 1931.85
2250 x cos20 = 2114.31

vertical=
2000 x sin15 = 517.64
2250 x sin20 = 769.55

total horizontal = 4046.16
total vertical = 1287.19

√4046.162+1287.192
= 4245.97

tan= 4046.16/1287.19
=3.14º

I'm pretty sure I messed everything up -_-
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KK.Violinist
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(Original post by KK.Violinist)
so that would be 4046.16N as resultant force. I'm not sure yet so I'll use the method given from the other post and get back to you
ok, that's the horizontal component.
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lerjj
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(Original post by KK.Violinist)
In order to try and recover a car stuck in a muddy field, two tractors pull on it. The first acts at an angle 20º left of the forwards direction with force of 2250N. The second acts 15º to the right of direction with force of 2000N. Find resultant force on the stuck car.

I understand it but not sure on how to solve it. Do I apply the parallelogram rule and add the two forces?

Thanks in advance.
That will work. As others have stated, another option is to add each components individually.

As an aside, I don't really like the parallelogram rule because it doesn't really help you to compute stuff - the 'triangle' picture seems clearer to me. (It's basically the same, but with one less construction line). You can then solve simply using the cosine rule in this case, although be careful when it comes to working out the angle.
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KK.Violinist
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(Original post by lerjj)
That will work. As others have stated, another option is to add each components individually.

As an aside, I don't really like the parallelogram rule because it doesn't really help you to compute stuff - the 'triangle' picture seems clearer to me. (It's basically the same, but with one less construction line). You can then solve simply using the cosine rule in this case, although be careful when it comes to working out the angle.
I thought the triangle rule only applies if vectors are head to tail, tail to head and parallelogram is for two vectors tail to tail but in same plane?
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KK.Violinist
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Can anyone correct me if I was wrong, would really appreciate it
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joostan
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(Original post by KK.Violinist)
Can anyone correct me if I was wrong, would really appreciate it
Not quite.
(Original post by KK.Violinist)
vertical=
2000 x sin15 = 517.64
2250 x sin20 = 769.55
One of these should be negative, as they act in opposite directions to each other. It doesn't matter which for the modulus, though if you want a positive angle, you may need to think about which angle you choose.
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KK.Violinist
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(Original post by joostan)
Not quite.

One of these should be negative, as they act in opposite directions to each other. It doesn't matter which for the modulus, though if you want a positive angle, you may need to think about which angle you choose.
Which angle to choose? Would I pick the greater of the two or...? I understand what you're saying up till the point about which angle to choose.
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joostan
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(Original post by KK.Violinist)
Which angle to choose? Would I pick the greater of the two or...? I understand what you're saying up till the point about which angle to choose.
Well, you can pick either angle (which is not the right angle) of the right angled triangle produced by the components of the force.
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KK.Violinist
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(Original post by joostan)
Well, you can pick either angle (which is not the right angle) of the right angled triangle produced by the components of the force.
ok, that makes more sense now, thanks. So the answer would be √4046.162+1287.192= 4245.97 as the resultant force and either of the angles as directions?
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joostan
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(Original post by KK.Violinist)
ok, that makes more sense now, thanks. So the answer would be √4046.162+1287.192= 4245.97 as the resultant force and either of the angles as directions?
Wait, the vertical component of one of the forces still needs to be negative. . .
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KK.Violinist
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(Original post by joostan)
Wait, the vertical component of one of the forces still needs to be negative. . .
oops.

horizontal=
2000 x cos15 = 1931.85
2250 x cos20 = 2114.31

vertical=
2000 x sin15 = 517.64 (-517.64)
2250 x sin20 = 769.55

total horizontal = 4046.16
total vertical = 251.91

√4046.162+251.912
=4053.99

any mistakes?
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joostan
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(Original post by KK.Violinist)
oops.

horizontal=
2000 x cos15 = 1931.85
2250 x cos20 = 2114.31

vertical=
2000 x sin15 = 517.64 (-517.64)
2250 x sin20 = 769.55

total horizontal = 4046.16
total vertical = 251.91

√4046.162+251.912
=4053.99

any mistakes?
Haven't checked the actual calculations- can't seem to find my calculator :'), but your reasoning is fine
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KK.Violinist
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(Original post by joostan)
Haven't checked the actual calculations- can't seem to find my calculator :'), but your reasoning is fine
what would I write as my answer (as a sentence). I usually lose marks explaining during exams so I want to make sure I fully understand it
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joostan
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(Original post by KK.Violinist)
what would I write as my answer (as a sentence). I usually lose marks explaining during exams so I want to make sure I fully understand it
Write the magnitude of the force, and then specify both the size of the angle, and the direction relative to either the horizontal or vertical of the resultant force.
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KK.Violinist
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(Original post by joostan)
Write the magnitude of the force, and then specify both the size of the angle, and the direction relative to either the horizontal or vertical of the resultant force.
okay, thank you very much! One last question haha, the tan=horizontal/vertical equation was to work out the angle of the triangle correct or no? If so, then I can just use that as my angle? Sorry if I sound dumb right now, haven't slept in awhile and am behind on physics after transferring schools
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joostan
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(Original post by KK.Violinist)
okay, thank you very much! One last question haha, the tan=horizontal/vertical equation was to work out the angle of the triangle correct or no? If so, then I can just use that as my angle? Sorry if I sound dumb right now, haven't slept in awhile and am behind on physics after transferring schools
No worries, and yes you can easily calculate \tan(\theta), you then simply need to take the arc-tangent, that is \tan^{-1} to find \theta.
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