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

Question

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 51.5 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 16.5 kg of food left for the rest of your trip so you put 16.5 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.

What will be the acceleration of the bags when you let go of the anchor bag in m/s^2?

Explanation / Answer

Your conclusion is fine --
a) anchor bag on slope ? food bag will drop, pulling the anchor bag up the slope

Draw free-body force diagrams on both masses (for the time being, allowing them to be different: m? = food mass; m? = anchor mass), and equate the acceleration of m? downward to that of m? up the slope, which makes an angle, ?, with the horizontal:

a = g - T/m? = T/m? - g sin?

Solve for T:
T = g(1 + sin?) / (1/m? + 1/m?)
a = g(1 - (1 + sin?) / (1 + m?/m?))

Given m? = m?, ? = 51.5, g = 9.81 m/s

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