Asked by Dan Ira King
                1.A block of masses mA is on a plane inclined at an angle θ with the horizontal. It is attached to another mass mB by means of string that passes over a pulley at the top of the incline. For mA = 8kg, mB = 5kg, θ = 20o, and the coefficient of kinetic friction between mA and the plane is 0.3. Calculate the (a) acceleration of the masses and (b) the tension in the string when the system is moving.
            
            
        Answers
                    Answered by
            Arora
            
    It would help with visualising if you drew block diagrams for the question.
Net force on B = downwards weight (mg) and upwards Tension (T)
Net force on A = down the incline weight (mgsinθ) and up the incline Tension (T)
I'm gonna take mass of A as M, and mass of B as m. The accelerations must be same, and are thus 'a' in both cases:
The equations formed are as follows:
1) T - Mgsinθ = Ma
2) mg - T = ma
Adding the two and eliminating T:
mg - Mg(sin20) = (M+m)a
Putting in the values:
5*10 - 8*10*0.34 = (8+5)a
=> a = (50-27.2)/13
= 22.8/13
= 1.75 m/sec^2
Now, for the Tension, you can simply put the value of a into one of the earlier equations, 1 or 2.
    
Net force on B = downwards weight (mg) and upwards Tension (T)
Net force on A = down the incline weight (mgsinθ) and up the incline Tension (T)
I'm gonna take mass of A as M, and mass of B as m. The accelerations must be same, and are thus 'a' in both cases:
The equations formed are as follows:
1) T - Mgsinθ = Ma
2) mg - T = ma
Adding the two and eliminating T:
mg - Mg(sin20) = (M+m)a
Putting in the values:
5*10 - 8*10*0.34 = (8+5)a
=> a = (50-27.2)/13
= 22.8/13
= 1.75 m/sec^2
Now, for the Tension, you can simply put the value of a into one of the earlier equations, 1 or 2.
                    Answered by
            Dan Ira King
            
    What happens to the coefficient of friction? Why is it not used?
    
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