Calculating Forces on Boy Sledging Down Slope

[raging_hippo]

A boy is traveling down a slope on a sledge. The boy and the sledge have a total mass of 60kg and are traveling at a constant speed. The angle of the slope to the horizontal is 35 degrees.
There are 3 forces acting on him, the weight of him and the sledge acting straight down W, the push from the slope acting perpendicular to the slope P and the resistive forces on the sledge and the boy acting up the slope.

Im completely stumped on how to do this, I've got to find the magnitude of the resistive force and determine the component of weight W that acts perpendiculay to the slope so if someone could answer the question and show how they did it i would be really grateful. Thanks

 

[deltabourne]

Have you drawn a force diagram? That makes it easy. You will then want to create a reference axis, which can be done parallel/perpendicular to the slope (the normal force and the resistive force already act on this axis). Since w acts straight down, we will want to break it into two components using trigonometry. You will find that you get w*sin(theta), parallel to the slope, and w*cos(theta), perpendicular to the slope, where theta is the angle between the slope and the x-axis (35 degrees in this case)

Next you want to write your net force equations. Starting with the parallel axis, we have:
F_net_parallel = w*sin(theta) - F_resistive = m*a, but since acceleration is 0, you have w*sin(theta0=F_resistive

Next, for the perpendicular axis:
F_net_perpendicular = w*cos(theta) - F_normal = m*a, and again a=0, so w*cos(theta) = F_normal

From there, all's that you have to do solve the equations for what you are looking for. Hope that helps.

 

[raging_hippo]

Cheers that's cleared it up nicely, thanks a lot