The Physics of Tobogganing Part 3

This is my last blog about sledding, I promise. Anyway, part 2 showed my uncle doing work to pull my cousin across the snow. However, as the pictures below illustrate, my brother also performed work to pull my cousin UP the hill of snow.

This classic example of laziness (on my cousin's part at least) demonstrate a number of physics concepts. The most obvious of course are Newton's Laws and work. As my brother pulls my cousin up the hill of snow, he does work and this is explained through forces. Because he's doing work up the incline, my brother has to exert a force greater than mgcos(theta) + F(friction). Since the surface is slanted, according to a FBD diagram, only a component of my cousin's weight mg affects the force that my brother has to exert. Additionally, due to the friction created between the sled and the snow which points opposite the direction of my brother's motion, the force he exerts must be greater thatn the combined values of mgcos(theta) and F(friction). We also have to take into account the friction between my brother's snow shoes and the snow as he moves up the hill. Thus, the force he exerts then must be greater than mgcos(theta), F(friction1), and F(friciton2). No wonder then that my sister had to lend a helping hand from the back. As my sister pushes my cousin forward, she accounts for some of the force value needed to pull my cousin back up the hill. In my sister's case, the force she exerts is similar to what my brother has to deal with. It also doesn't help that my cousin does nothing to propel himself forward or lean forward in order to lessen the force that he's exerting towards the back of the sled and therefore opposite of the direction desired. Because he's leaning back, he is exerting a force of his own that adds to the mgcos(theta) already established.

As my brother and sister work together to pull my cousin back up the hill of snow, they do work. As in the previous blog, because my brother and sister are exerting force at an angle, the value of the work they perform is greater that it would have been if they were pulling parallel to the sled's orientation. At this point, the value of cos(theta) would just been 1 and the work equation would have been the simple W=F(net)x.