Finding Accerleration on a slope with friction

[fallen186]

**Sorry the latex didn't really work for me.**

Homework Statement

In the figure below, m1 = 3.6 kg, m2 = 5.1 kg, and the coefficient of kinetic friction between the inclined plane and the 3.6-kg block is μk = 0.25. Find the magnitude of the acceleration of the masses.

m1 = 3.6 kg
m2 = 5.1 kg
μ = 0.25
Theta = 30

Homework Equations

a = ((m2*g)-(m1*g*sin(theta)))/(m1+m2)

The Attempt at a Solution

Since there is friction I did $$(m1*g*sin(theta)-μ*Fn <br /> Fn = (.5*m1*g*3^.5)<br /> <br /> a = ((5.1 kg*9.8 m/s^2) - (3.6 kg*9.8 m/s^2*sin(30)-.25(.5*3.6kg*9.8m/s^2*3^.5)))/(3.6kg+5.1kg)<br /> a = (50.031- (17.658 - 7.638))/(8.7)<br /> <br /> a = 4.599 m/s^2<br /> <br /> I got this wrong. please help$$

 

[alphysicist]

Hi fallen186,

**Sorry the latex didn't really work for me.**

Homework Statement

In the figure below, m1 = 3.6 kg, m2 = 5.1 kg, and the coefficient of kinetic friction between the inclined plane and the 3.6-kg block is μk = 0.25. Find the magnitude of the acceleration of the masses.

m1 = 3.6 kg
m2 = 5.1 kg
μ = 0.25
Theta = 30

Homework Equations

a = ((m2*g)-(m1*g*sin(theta)))/(m1+m2)

The Attempt at a Solution

Since there is friction I did $$(m1*g*sin(theta)-μ*Fn$$

$$<br /> There's not much detail about the problem, but the minus sign here means that m1 is moving down the incline. Is that what you intended?$$