Here's a classic question for the end of the equilibrium unit. I say "classic" because although I got the picture from Giancoli's text, I first encountered the question in Dave Ledden's physics class when I was a senior in high school.
A bear sling, as shown above, is used in some national parks for placing backpackers’ food out of reach of the federal bears. Is it possible to pull the rope hard enough so that it doesn’t sag at all? (Obviously, justify your answer in a couple of sentences… just “yes” or “no” doesn’t cut it :-) )
A good explanation points out that with no sag, no upward force would counteract the bag's weight; thus the bag would not be in equilibrium, and the bag must fall. An even better explanation would explain that the vertical components of the rope's tension provide the upward force to counteract the bag's weight; in order for vertical components of tension to exist, the rope must pull somewhat upward, and so must sag.
An outstanding explanation shows that the vertical components are each Tsinθ, where θ is the angle of the rope measured from the horizontal. If the rope approaches purely horizontal, the angle goes to zero. The vertical equilibrium statement is 2(Tsinθ) = mg. As θ goes to zero, sin θ also goes to zero... meaning that the equilibrium statement for a horizontal rope is mg=0. Impossible!
Another equally appealing explanation (as you know, but might have thought to keep aside) would be as follows:
ReplyDeleteTo keep the rope horizontal and to produce an upward balancing force (which is the resultant of the tensions), the tension T has to be infinite, which is impossible.
With best wishes,
MV
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ReplyDeleteis there an answer for representing forces
ReplyDeleteKhaled and king, take a look at the last paragraph... it gives the best sort of answer that I would expect from my students.
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