ma hiker in Glacier National Park, you are looking for a way to keep the bears f

Question

ma 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 fall 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 14.5 kg of food left for the rest of your trip so you put 14.5 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. The motion of the food bag cannot be determined. The anchor bag is on a slope so the food bag will drop, pulling the anchor bag up. Nothing. The bags have the same mass so they will not move. The anchor bag is on a slope so it will pull the food bag back up. 55.5° What will be the acceleration of the bags when you let go of the anchor bag? Number m/s²

Explanation / Answer

The anchor bag is on the slope so the food bag will drop , pulling the anchor bag up

Writing the force equations we have

T - mg sin 55.5 = ma --------(1)

mg - T = ma ---------(2)

Adding both equations we ahve

14.5 *9.8*(1- sin 55.5) = 2*14.5*a

a = 0.86 m/s2

so the acceleration of the bags = 0.86 m/s2

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