A) 39N
A (30, 60, 90 deg.) triangle of forces is involved here. The principle is: when three forces are in equilibrium at a point, their ratios are the same as those of the corresponding lengths in a triangle formed from their three directions.
I'll work in kg-equivalent-masses first, converting to N (Newtons) only at the end.
The kg-equivalent forces are those of the weight due to a mass of 2.0 kg acting downwards (short "perpendicular side" of triangle), of
2 sqrt(3) kg horizontally (long perp. side) and of 4.0 kg along the hypoteneuse, normal to the board. (The ratios of these sides, 1:sqrt(3):2 are the ratios for the 30, 60, 90 deg. triangle.)
So the normal force exerted by the book is the force due to the weight of a 4.0 kg mass, that is 4.0 x 9.8 N =
39 N.
(4.0 x 9.8 = 39.2, but only 2 sig. figs. are justified by the input data.)
Live long and prosper.
POSTSCRIPT: WARNING!! pd's link, below, answers a DIFFERENT question, in which FRICTION is involved. That problem specifies a value for the coefficient of friction. That is actually a slightly more complicated problem.
The statement of your problem doesn't give any coefficient of friction, so I assumed that the board is frictionless. (The problem as set is frankly somewhat deficient in not making the question of friction or no friction explicit. A well-set problem would make the situation clear.) Nevertheless, as you can see, equilibrium IS possible under just the three forces of (i) the weight of the book, (ii) the horizontal force "pinning it in place," and (iii) the normal reaction that the (frictionless) board produces. AND, the answer is one of the 4 choices you were given! --- always a good sign.
(When friction enters a problem like this, you can view it either as a 4-force problem, or alternatively a 3-force problem in which the board's reaction is NOT JUST NORMAL to it.)
2006-12-30 13:15:34
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answer #1
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answered by Dr Spock 6
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2016-10-19 06:15:57
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answer #2
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answered by ? 4
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