prob_062 - 62. REASONING The weight of the part of the...

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Unformatted text preview: 62. REASONING The weight of the part of the washcloth off the table is moff g. At the instant just before the washcloth begins to slide, this weight is supported by a force that has magnitude equal to fsMAX, which is the static frictional force that the table surface applies to the part of the washcloth on the table. This force is transmitted “around the bend” in the washcloth hanging over the edge by the tension forces between the molecules of the washcloth, in much the same way that a force applied to one end of a rope is transmitted along the rope as it passes around a pulley. SOLUTION Since the static frictional supports the weight of the washcloth off the table, we have fsMAX = moff g. The static frictional force is fsMAX = µsFN . The normal force FN is applied by the table to the part of the washcloth on the table and has a magnitude equal to the weight of that part of the washcloth. This is so, because the table is assumed to be horizontal and the part of the washcloth on it does not accelerate in the vertical direction. Thus, we have f sMAX = µ s FN = µ s mon g = moff g The magnitude g of the acceleration due to gravity can be eliminated algebraically from this result, giving µsmon = moff . Dividing both sides by mon + moff gives µs ⎜ ⎛ ⎜ mon + moff ⎝ mon ⎞ moff ⎟= ⎟ mon + moff ⎠ or µ s f on = f off where we have used fon and foff to denote the fractions of the washcloth on and off the table, respectively. Since fon + foff = 1, we can write the above equation on the left as µ s 1 – f off = f off ( ) or f off = µs 0.40 = = 0.29 1 + µ s 1 + 0.40 ______________________________________________________________________________________ ______ ...
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