Forces on an inclined plane, kinetic friction How is this wrong?

[neutron star]

Homework Statement

In the figure below, m1 = 3.8 kg, m2 = 5.5 kg, and the coefficient of kinetic friction between the inclined plane and the 3.8-kg block is μk = 0.26. Find the magnitude of the acceleration of the masses and the tension in the cord.

http://img29.imageshack.us/img29/2676/0541.gif

Homework Equations

The Attempt at a Solution

5.5kg*10=55N
38 sin(30)=19N
(19N)(0.26)=5N

55-T=5.5a
+ T-19-5=3.8a
_______________
55-19-5=9.3a
39=9.3a
(divide both sides by 9.3)
a=3.333
(plug 3.333 in for a)
55-T=5.5(3.333)
55-T=18.33
-T=-36.67
T=36.67

a=3.333
T=36.67

How is this wrong? It says I am incorrect, why?

a =___ m/s2
T = ___ N

 

[Rock32]

Homework Statement

In the figure below, m1 = 3.8 kg, m2 = 5.5 kg, and the coefficient of kinetic friction between the inclined plane and the 3.8-kg block is μk = 0.26. Find the magnitude of the acceleration of the masses and the tension in the cord.

http://img29.imageshack.us/img29/2676/0541.gif

I would personally advise keeping the equations in the form of algebraic variables (m1/m2/g/a/T), then solving for the variables. When you get to more complicated problems, doing all of the necessary algebra before plugging in numbers makes it easier not to make mistakes.

The Attempt at a Solution

5.5kg*10=55N
38 sin(30)=19N
(19N)(0.26)=5N

55-T=5.5a
+ T-19-5=3.8a

You're rushing things a bit here. Make both x and y direction equations for m1. Make the +x direction along the plane of the slope going towards the pulley, and the y direction perpendicular to the slope.

The y direction equation for the box is:

N (normal force) - m1gcos(a) = 0
N = m1gcos(a)

The x direction for box m1 is:

T - m1gsin(a)-(μk)(N) = m1a1

For box m2:

m2g - T = m2a2


Now try combining, and then solve. (For this problem, assume a1 = a2, and the Tension is the same everywhere)

 

[neutron star]

I would personally advise keeping the equations in the form of algebraic variables (m1/m2/g/a/T), then solving for the variables. When you get to more complicated problems, doing all of the necessary algebra before plugging in numbers makes it easier not to make mistakes.



You're rushing things a bit here. Make both x and y direction equations for m1. Make the +x direction along the plane of the slope going towards the pulley, and the y direction perpendicular to the slope.

The y direction equation for the box is:

N (normal force) - m1gcos(a) = 0
N = m1gcos(a)

The x direction for box m1 is:

T - m1gsin(a)-(μk)(N) = m1a1

For box m2:

m2g - T = m2a2


Now try combining, and then solve. (For this problem, assume a1 = a2, and the Tension is the same everywhere)

I don't think it's possible to solve that without having a.

You have:

T-38sin(a)-(0.26)(N)=3.8a
+ 55-T=5.5a

It would be 38sin(a)-(0.26)(N)+55=9.3a
You can't put in the normal force because you need acceleration to solve m1gcos(a) and you can't get rid of that 38sin(a). I'm guessing you meant 38sin(30).

 

[Rock32]

I don't think it's possible to solve that without having a.

You have:

T-38sin(a)-(0.26)(N)=3.8a
+ 55-T=5.5a

It would be 38sin(a)-(0.26)(N)+55=9.3a
You can't put in the normal force because you need acceleration to solve m1gcos(a) and you can't get rid of that 38sin(a). I'm guessing you meant 38sin(30).

Yes, sorry for not clarifying. The (a) inside the trig functions stood for the angle (30), and a1/a2 were the acelerations.