Question

In the figure below, m1 = 3.2 kg, m2 = 5.3 kg, and the coefficient of kinetic friction between the inclined plane and the 3.2-kg block is μk = 0.26. Find the magnitude of the acceleration of the masses and the tension in the cord.

the ramp is 30 degrees m2 is hanging

Answer 1

let's start with m2:

Fnet = m₂a = 5.3 kg *a = W - T₂

5.3a = m₂g - T₂  

5.3a = 5.3 * 9.8m/s² - T₂

(That is, I assume that m2 will fall, which means that its weight exceeds the tension in the cord. I'll call this "positive" acceleration.) So
T₂ = 51.9 - 5.3a

Now for m1:
gravity parallel to plane: Fg = m₁gsinΘ = 3.2kg * 9.8m/s² * sin30 = 15.7 N
friction: F_f = µm₁gcosΘ = 0.26 * 3.2kg * 9.8m/s² * cos30 = 7.06 N
tension: T₁
Fnet = m₁a = 3.2kg * a = T₁ - Ff - Fg = T₁ - 15.7N - 7.06 N = T₁ - 22.8 N
(That is, the net acceleration on m1 is the tension pulling it up the slope less friction and gravity.) So
T₁ = 3.2a + 22.8 N

Since T₂ = T₁,
51.9 - 5.3a = 3.2a + 22.8
29.1 = 8.5a
a = 3.42 m/s² ← acceleration

Tension in cord, T₂ = 51.9 - 5.3a = 51.9N - 5.3kg * 3.42m/s² = 33.77 N ←
Check:
T₁ = 3.2a + 22.8 = 3.2kg * 3.42 m/s² + 22.8 N = 33.74 N