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prosteve037
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Homework Statement
I've already answered this question correctly after multiple tries but I am still confused about the method in solving these kind of problems.
The two blocks (m = 18 kg and M = 88 kg) in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is μs = 0.33, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force [itex]\vec{F}[/itex] required to keep the smaller block from slipping down the larger block?
http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c06/fig06_36.gif
Homework Equations
Newton's Second Law: [itex]\textit{ƩF = ma}[/itex]
Newton's Third Law: [itex]\textit{F}_{A}\textit{ + F}_{B}\textit{ = 0}[/itex]
The Attempt at a Solution
For problems like these, why and how is it that the total acceleration is equal to the individual accelerations of the bodies?
Is it a crucial step in solving problems like these to find the net force and net/total acceleration of the two bodies?
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There was another question I read that showed the separate coefficients of friction (between the bodies and the ground) for two bodies on a surface (interacting/touching each other). Given one of the interacting forces between the two objects, would the net/total acceleration of the system still be the individual accelerations of the bodies?
Here's an example of this kind of problem that I'm talking about:
In the figure here, a box of Cheerios (mass mC = 1.4 kg) and a box of Wheaties (mass mW = 3.1 kg) are accelerated across a horizontal surface by a horizontal force applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.6 N, and the magnitude of the frictional force on the Wheaties box is 4.8 N. If the magnitude of [itex]\vec{F}[/itex] is 12.2 N, what is the magnitude of the force on the Wheaties box from the Cheerios box?
http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c06/fig_6_A.gif
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