10/13/2014 01:55 PM Tol 100 010/13/201401:51 PM sapling learning As a National y
ID: 1282893 • Letter: 1
Question
10/13/2014 01:55 PM Tol 100 010/13/201401:51 PM sapling learning As a National you looking for a way to keep the bears from getting at your supply You find a cam d that near of ice from Part of the outcropping forms a 55.5 slope up to a vertical cliff, You 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 19.5 of foad left for the rest of your trip so you put 19,5 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 to be frictionless The anchor bag is on a slope so the food drop, pulling the anchor bag o The motion of the food bag cannot be determined. o The anchor bag is on a slope so it will pull the food bag back up. o Nothing. The bags have the same mass so they will not move. 55.50 What will be the acceleration of the bags when you let go of the anchor bag? m/s 23.661425 View Previous view Next SExi TOSHIBAExplanation / Answer
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?, ? = 55.5
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.