A 29.5-kg block (m1) is on a horizontal surface, connected to a 5.5-kg block (m2

Question

A 29.5-kg block (m1) is on a horizontal surface, connected to a 5.5-kg block (m2) by a massless string as shown in the Figure. The pulley is massless and frictionless. A force of 196.1 N acts on m1 at an angle of 29.9o. The coefficient of kinetic friction between m1 and the surface is 0.209. Determine the upward acceleration of m2. Here is a link to the figure that should work http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype10/prob06_horzblock.gif Please help, I can't figure out this problem and my teacher doesn't do any math problems AT ALL in class its all conceptual.

Explanation / Answer

there may error in calculation but method is correct

 

 

The main thing to keep in mind about these problems is that all blocks in the system accelerate at the same rate (at least in my experience).
Let's start with the vertical forces on m1:
Fv = mg - FsinΘ = 29.5kg · 9.8m/s² - 219.3N·sin29.9º = 283N - 196.1N = 173N
So the force of friction is
Ff = µFn = 0.209· 173N = 33.4N

free-body diagram on m1 (horizontal forces only):
Fh = ma = 29.5kg · a = FcosΘ - (Ff + T) = 196.1N·cos29.9º - 33.4N - T = 156.3N - T
where T is the tension in the string. So
T = 156.3N - 28.9kg · a

fbd on m2 (vertical only)
Fv = ma = T - mg
So T = ma + mg = 5.5kg · a + 5.5kg · 9.8m/s² = 5.5kg · a + 53.9 N

Since T = T (that is, the tension in the string is constant throughout),
5.5kg · a + 53.9 N = 156.3N - 28.9kg · a
34.4kg · a = 102.4N
a = 2.98 m/s² ← ans

Check: see if forces balance -- I'll check m2 first:
T = 5.5kg · a + 53.9 N = 5.5kg · 2.98m/s² + 53.9 N = 70.3N

now m1:
T = 156.3N - 29.5kg · a = 156.3N - 28.9kg · 2.98m/s² = 70.2 rounding err √

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