http://imgur.com/qePp1Rs You are lowering two boxes, one on top of the other, do
ID: 1278762 • Letter: H
Question
http://imgur.com/qePp1Rs
You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 18.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.428, and the coefficient of static friction between the two boxes is 0.785.
What force do you need to exert to accomplish this?
What is the magnitude of the friction force on the upper box?
I found similar questions to this on yahoo answers and chegg as well but when i follow the equations with my own numbers its telling me I'm wrong..
Explanation / Answer
then A)
the net force needed is F = uk ( m1 + m2 ) g + us m2 g ,
where 'm1' is mass of lower box, and 'm'2 is that of the upper box.further 'us' is the coeffficient of static friction and 'uk' is the coefficient of kinetic friction.
F = 0.428 * (32+48) * 9.81 + 0.785 * 32 * 9.81
= 335.8944 + 246.4272
= 582.3216 N
B) the friction on the upper box is fs = us m2 g and its direction will be opposite to that of the applied force.
fs = 0.785 * 32.9.81
fs = 246.4272 N = 246.43 N
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