Newton's Law Problem: Calculating Acceleration on an Inclined Plane

[Smile101]

Homework Statement

A block of mass m1=2.0 kg on a frictionless inclined plane of angle 20 degrees is connected by a rope over a pulley to another block of mass m2=1.0kg. What are the magnitude and direction of the acceleration of the second block?

Homework Equations

a=m2g-m1gsin(angle)/ma1+m2

The Attempt at a Solution

how would i calculate m1g and m2g in the first place?
other then that...

a=m2g-m1gsin20/2+1



Your help will beeee very appreciated seeing as I have a test on it tom!

 

[rock.freak667]

g is acceleration due to gravity = 9.81 m/s²

 

[Smile101]

so basically i would multiply mass with 9.81
m2 = 1(9.81) = 9.81
m1= 2(9.81) = 19.62

a= -19.62-(-9.81)sin20/3
a= 3.3m/s^2 (down)

but the answer is supposed to be -1.03 m/s^2 (down)

 

[rock.freak667]

so basically i would multiply mass with 9.81
m2 = 1(9.81) = 9.81
m1= 2(9.81) = 19.62

a= -19.62-(-9.81)sin20/3
a= 3.3m/s^2 (down)

but the answer is supposed to be -1.03 m/s^2 (down)

Well for one, it seems that you switched around m₁g and m₂g. It should be

$$a=\frac{9.81-(19.62)sin20}{3}$$


Also, do you know how to get that formula?

$$a=\frac{m_2g-m_1gsin\theta}{m_1+m_2}$$

 

[Smile101]

Well for one, it seems that you switched around m₁g and m₂g. It should be

$$a=\frac{9.81-(19.62)sin20}{3}$$


Also, do you know how to get that formula?

$$a=\frac{m_2g-m_1gsin\theta}{m_1+m_2}$$

No i don't! But the modified version you gave me is right! I got 1.03! :) thank you soo much

 

[rock.freak667]

No i don't! But the modified version you gave me is right! I got 1.03! :) thank you soo much

If the resultant force of m2 is downwards, then for the system vertically. (T=Tension)

m²a=m₂g-T

therefore on the incline, m₁ moves up, so that

m₁a=T-m₁gsin$\theta$


two equations where you can eliminate T by adding them. That is how to derive that formula. It's better to know how to do these kinds of of problems from first principles than to memorize a formula