M*g = 21.5 * 9.8 = 210.7 N. = Wt. of food.
Fpd = 210.7*sin55.5 = 173.6 N. = Force parallel to the incline and downward.
sin55.5 = 210.7/Fpu. Fpu = 256 N. = Force parallel to the incline and upward.
Since the force acting up the incline is greater than that acting down the incline, the food will fall to the ground.
As a hiker in Glacier National Park, you are looking for a way to keep the bears from getting at your supply of food. You find a campground that is near an outcropping of ice from one of the glaciers. Part of the ice outcropping forms a 55.5° slope up to a vertical cliff. You decide that this is an ideal place to hang your food supply as the cliff is too tall for a bear to reach it. You put all of your food into a burlap sack, tie an unstretchable rope to the sack, and tie another bag full of rocks to the other end of the rope to act as an anchor. You currently have 21.5 kg of food left for the rest of your trip so you put 21.5 kg of rocks in the anchor bag to balance it out. What happens when you lower the food bag over the edge and let go of the anchor bag? The weight of the bags and the rope are negligible. The ice is smooth enough to be considered frictionless.
1 answer
1 answer