Answer:
A) The resultant force is 43.4 [N]
B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis
Explanation:
We need to make a sketch of the different forces acting on the heavy crate.
In the attached image we can see the forces and the sum of the vector with their respective angles.
Forces in the X-axis

Forces in the y-axis
![FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]](https://tex.z-dn.net/?f=FDiony%3D0%5BN%5D%5C%5CFshirley%3D%2016.5%2Asin%2830%29%3D8.25%5BN%5D%5C%5CFjoany%3D19.5%2Asin%2860%29%3D16.88%20%5BN%5D%5C%5C%5C%5CForcesy%3D0%2B8.25-16.88%3D%20-8.63%5BN%5D)
Using the Pythagorean theorem

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:
