Physics friction problem(advice/tips is greatly appreciated)?

A string going over a massless frictionless pul-


ley connects two blocks of masses 4.1 kg and


7.9 kg. As shown on the picture below, the


4.1 kg block lies on a 31 鈼?incline; the coeffi-


cient of kinetic friction between the block and


the incline is 渭 = 0.32. The 7.9 kg block is


hanging in the air.


The 7.9 kg block accelerates downward


while the 4.1 kg block goes up the incline


with the same acceleration.








G=9.8m/s^2 What is the acceleration?





Any formula or advice on formula is thankedPhysics friction problem(advice/tips is greatly appreciated)?
Where is the illustration of the system ';As shown on the picture below?';





Anyway, in the absence of the picture, I think I can still construct the free body diagrams of the two masses.





For the 4.1-kg mass,





Equation 1 is





T - f - (M1)g(sin 31) = (M1)a





where





T = tension in the string pulling the 4.1-kg mass


f = frictional force opposing the motion of the 4.1-kg mass


M1 = 4.1 kg (given)


g = acceleration due to gravity = 9.8 m/sec^2 (constant)


a = acceleration





By definition,





frictional force = f = (mu) * (Normal force)





From the free body diagram,





Normal force = (M1)(g)(cos 31)





Therefore, Equation 1 modifies to





T - (mu)M1(g)((cos 31) - M1(g)(sin 30) = (M1)a





and solving for ';T';,





T = (M1)a +(mu)(M1)(g)((cos 31) + M1(g)(sin 30)








For the 7.9-kg block, the working equation is





M2(g) - T = (M2)(a)





where





M2 = 7.9


g = 9.8


T = tension in the string holding the 7.9-kg block


a = acceleration





Solving for ';T';





T = M2(g) - M2(a)





and since the pulley is frictionless, then equating the two ';T's';





(M1)a + (mu)(M1)(g)((cos 31) + M1(g)(sin 30) = M2(g) - M2(a)





Combining terms





M1(a) + M2(a) = M2(g) - [(mu)(M1)(g)((cos 31) + M1(g)(sin 30)]





a(M1 + M2) = g[(M2) - (mu)(M1)(cos 31) - (M1)(sin 31)]





Substituting appropriate values,





a(4.1 + 7.9) = 9.8[(7.9 - (0.32)((4.1)(cos 31) - (4.1)(sin 31)]





12(a) = 9.8[7.9 - 1.12 - 2.11] = 9.8(4.67)





12(a) = 45.77





a = 3.81 m/sec^2











Physics friction problem(advice/tips is greatly appreciated)?
We kinda need a picture to help...