Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25.0° west of north, and the resultant of these two pulls is 460.0 N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

Answer : Here there are two forces One force is small (F) and the other force is two times bigger or greater than (F) i.e 2F Now here , The north component gets added to 460 N. We can write the component vectors as follows Cosine component: 2f Cos 25 + Fcos = 460N ……………..(1) Sine component = 2f sin 25 - fsin = 0 …………….(2) Where is measured form Y- axis Now from equation (2) , F gets cancelled, Therefore equation (2) becomes 2sin 25 = sin = .845 as = 58 ( East to north) Using equation 1 , we get 2f Cos 25 + Fcos 58 = 460N = 1.813 + 0.534F = 2.347F = 460 N F= 196N 2F= 392 N Hence the magnitude of each of these pulls F= 196N , 2F= 392 N,