Kalle and Ville want to move the picture box with two ropes on the same level. If Kalle pulls the box with a force of 550 N, then with how much force does Ville has to apply to the rope to pull it so that the box moves along the dotted line shown in the picture.
I tried calculating this in the following way: sin(25.8/550)=sin(34.9/y) to have the forces in equilibrium but somehow this does not seem reasonable to me. I have had this problem before where I had to compare the forces given angles and which angle provided larger force compared to the other, but I do not know if same logic can be used here?
I also tried thinking this as of parallelogram where the angles are 60,7 x2 and 2x 119,3 respectively and the other sides being 550.
Am I looking at this problem the wrong way? The box has no assigned weight, friction coefficient or anything.
You want the component of the force that is perpendicular to the dotted line to cancel out. This component is given by $550\sin(25.8^\circ)$ N for the force that Kalle applies.
The component of force that Ville applies perpendicular to the dotted line is $y\sin(34.9^\circ)$ N, so we need $y\sin(34.9^\circ)=550\sin(25.8^\circ)$ which gives $y=\frac{550\sin(25.8^\circ)}{\sin(34.9^\circ)}$ N.
For me, it always helps to think of the units that numbers have. So instead of just "$34.9$", I write $34.9^\circ$, and similarly, write $550 N$ instead of $550$. Now remember that $\sin$ is a function that should receive degrees, so putting $550$ N in a sine function is not good.