Question

There are two forces on the 2.02 kg box in the overhead view of the figure but only one is shown. For F1 = 15.0 N, a = 13.4 m/s2, and θ = 22.2°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)

Here is the picture: http://ctrlv.in/237019

I don't understand where the second force is supposed to be. The question says that it isn't shown, so I don't understand if it is the sum of F1 and the force from a, or something else. Thanks!

Answer 1

its ur job to find F2.. :D :D

sum of F1 and F2 will give you an acceleration in the given direction!

Answer 2

I did that, and all of my answers were wrong again. I set (ma)cosθ=F1cos(0)+F2 to find F2 in the x direction, and then the same thing with sinθ for the y direction. Those were my answers for the unit vector notation.

For the magnitude, I took the square root of the sum of the x and y components of F2 (both values were squared).

I don't understand what I'm missing here.

Answer 3

second force =15/sin(22.2)

Answer 4

is it 39.69?

Answer 5

-39.69i

Answer 6

Is that the answer for the magnitude of the second force, or the value for the y component for the second force? Also how did you get that answer?

Answer 7

it is the value of second force

Answer 8

which is acting in - direction

Answer 9

by using this value u can easily calculate the force which is acting at 22.2 degree.