What Force is Needed for 1.5 m/s² Acceleration in the +y Direction?

[DaOneEnOnly]

Homework Statement

Two forces act on an object of mass 2.5kg; force F1 that is directed alont the +x-direction and has magnitude of .50N and force F2 that points at a 45 degree angle in the +y and -x quadrant and has magnitude 2.0N. Find the additional force, if any such that the object will accelerate in the +y direction with magnitude 1.5m/s/s.

Homework Equations

F=ma
Fnet=F1+F2+F3+etc...
F=Fsin$$\theta$$

The Attempt at a Solution

F_nety= F₂sin$$\theta$$+F_X-F_g

F_X= missing force= ?
F_nety= (2.5kg)(1.5m/s/s)= 3.75N
F₂sin$$\theta$$= 1.414N
F_g= 2.5*9.8=24.5N

3.75 = F_X + 1.414 -24.5
F_X= 26.84N

I'm not sure if I'm even trying to go about the problem correctly.

The selected answers in the back of the book has the answer as:
(0.91N)i+(2.3N)j

so I'm really confused.

 

[rl.bhat]

Let the third force be xi + yj.
Take x and y components of all the three forces. Since the resultant force is along y axis, total x-component must be zero.
Hence total y component must be equal to ...?

 

[tomcue]

Your attempt at a solution is on the right track, but there are a few errors. Firstly, the equation F=ma should be used to find the net force in the y-direction, not the x-direction. Secondly, the equation F=Fsinθ is only applicable when the force is acting at an angle to the horizontal, which is not the case for F1. Lastly, you have used the value for F1 in your calculation for F2sinθ, which is incorrect.

The correct approach would be to first find the net force in the y-direction:
Fnet_y = F2sin45° - Fg = 1.414N - 24.5N = -23.086N

Then, use the equation F=ma to solve for the missing force:
F = ma
F = (2.5kg)(1.5m/s^2)
F = 3.75N

Finally, use the Pythagorean theorem to find the magnitude of the missing force:
F = √(Fx^2 + Fnet_y^2)
3.75N = √(Fx^2 + (-23.086N)^2)
Fx = √(3.75N^2 - (-23.086N)^2)
Fx = 22.961N

Therefore, the missing force in the x-direction is 22.961N. To find the direction of this force, we can use the inverse tangent function:
θ = tan^-1(Fnet_y/Fx)
θ = tan^-1(-23.086N/22.961N)
θ = -45°

So the force Fx = 22.961N is acting at a 45° angle in the -x and -y quadrant. Therefore, the additional force needed to accelerate the object in the +y direction with a magnitude of 1.5m/s^2 is:
F = (22.961N)cos(-45°)i + (22.961N)sin(-45°)j
F = (16.24N)i + (-16.24N)j
F = (0.91N)i + (2.3N)j

This matches the answer given in the back of the book.