(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

There are two forces on the 3.2 kg box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the box. F_{1}is in Q I and F_{2}is in Q III

a = 12m/s^{2}

there is a picture here that I do not know how to upload -- it shows F_{1}in the +x direction ( 0º ) and F_{2}in Q III in the -x and -y direction at 30º from the y axis in the SW direction

then below that it gives the a=12m/s_{2}

I am not sure if that means a=12m/s_{2}for F_{2}or if that is for the f_{net}= ma where that is a......

(a) Find the second force in unit-vector notation.

(b) Find the second force as a magnitude and direction.

2. Relevant equationsand3. The attempt at a solution

(a) Find the second force in unit-vector notation.

F_{net}= ma

so F_{net}= 3.2kg * 12m/s^{2}= 38.4N

F_{2}= 38.4N - F_{1}= 38N - 20N = 18.4

(b) Find the second force as a magnitude and direction.

F_{2}=ma - F_{1}= [3.2kg * 12m/s^{2}]-20N = 18.4N

F_{2x}= 18 cos 240º = -9_{i}

F_{2y}= 18 sin 240º = -15.59_{i}

I am supposed to be getting -39.2_{i}and -33.3_{j}for F_{2}for part A and 51.4N (-140)º for part b. if they gave me ø how am I supposed to get -140º for ø? I am very confused with this question.

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# Vector addition

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