A sled is pulled forward by two dogs, each attached to the sled by a tow line. The angle between the two tow lines is $35^\circ$. How may times harder must one dog be pulling if the sled's path is at an angle of $20 ^\circ$ to the left hand dog's towline?
I decided to use the sine law:
$${r \over \sin (145)} = {b_1 \over \sin(17.5)} \\ r = {b_1 \sin (145) \over \sin (17.5)}$$
and for the second diagram:
$${r \over \sin (145)} = {b_2 \over \sin(20)} \\ r = {b_1 \sin (145) \over \sin (20)}$$
and set two equations equal to each other:
$$ {b_1 \sin (145) \over \sin (17.5)} = {b_2 \sin (145) \over \sin (20)} \\ 1.9074 b_1 = 1.6770b_2 \\ 1.137 b_1 = b_2$$
Therefore, one dog must pull 1.137 times harder.
However the answer is supposed to be 1.32 times harder.
What did I do wrong? I assume that it is wrong to assume that the resultant vector is the same. But I cannot find a way to solve this problem.
Thank you
If the path is horizontal, the force exerted by the first dog is $$(F_1\cos20^\circ,F_1\sin20^\circ)$$ and that by the second is $$(F_2\cos15^\circ,-F_2\sin15^\circ).$$ The sum of the forces must be horizontal, so $$F_1\sin20^\circ=F_2\sin15^\circ$$ so you can work out $F_2/F_1$.