Aboy pulls a sled with a force of 40n on a rope 2.5 meters long. the end that he holds is 1.5 meters higher than the end attached to the sled. what is the magnitude of the horizontal component that acts to pull the sled forward? first find the angel of the 40-n force.

Since the rope is 2.5m long and it is 1.5 m above the end attached to the sledge hence

sin (theta) = 1.5 / 2.5 so that theta is 36.9 degrees or rounding it off to 37degrees

You can see for yourself that this must be true as the rope is slanted with respect to the horizontal direction and so must represent the hypotenuse and the perpendicular distance from the boys hand to the horizontal axis is 1.5m. In this case take the origin of the coordinate system at the point where the rope and the sledge are attached. I think now things are pretty clear as far as the angle in concerned.

Now with respect to this origin the horizontal component of the force is give by

Fx = F cos(theta) in magnitude
= 40newtons * cos(37degrees)
= 31.9newtons which is about 32newtons when rounded to 2 significant figures

Ithink they do have the same properties just in less quantities i think