Physics AS question (resultant forces/vectors)

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.

Could you do 2250 x cos20 + 2000 x cos15 ?

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 $\theta$

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

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.

Can anyone correct me if I was wrong, would really appreciate it

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

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.

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?

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?

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

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