What is the Correct Calculation for the Second Force in Vector Notation?

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ganondorf29
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Homework Statement


Suppose that the 1 kg standard body accelerates at 3.40 m/s2 at 155° from the positive direction of the x axis, owing to two forces, one of which is F1 = (2.6 N)i + (4.8 N)j. What is the other force in unit-vector notation and as a magnitude and direction? i-component of the other force?

Homework Equations



F=ma

The Attempt at a Solution



First I tried to find the x-component of F2.

F2,x = max - F1,x
F2,x = (1)(3.4 cos155) - (2.6 cos 155)
I got F2,x to be -0.725 N but that's not the correct answer

For F2,y I did:
F2,y = may - F1,y
F2,y = (1)(3.4sin 155) - (4.8sin 155)
But its also wrong. Any ideas on what I am doing wrong?
 
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ganondorf29 said:

Homework Statement


Suppose that the 1 kg standard body accelerates at 3.40 m/s2 at 155° from the positive direction of the x axis, owing to two forces, one of which is F1 = (2.6 N)i + (4.8 N)j. What is the other force in unit-vector notation and as a magnitude and direction? i-component of the other force?

Homework Equations



F=ma

The Attempt at a Solution



First I tried to find the x-component of F2.

F2,x = max - F1,x
F2,x = (1)(3.4 cos155) - (2.6 cos 155)
I got F2,x to be -0.725 N but that's not the correct answer

For F2,y I did:
F2,y = may - F1,y
F2,y = (1)(3.4sin 155) - (4.8sin 155)
But its also wrong. Any ideas on what I am doing wrong?

I would suggest you try converting the 3.4 to i, j notation.

Now they give you one of the Forces F1 and the resulting force F3 (the 3.4N force), where

F3 = F1 + F2

What they are asking for is F2 which can be given by

F2 = F3 +(- F1)
 
Thank you. I got it right