Inclined Plane w/ Two Masses: F=ma+mb?

[ksm2288]

Homework Statement

Block of mass mA = 2kg - lying on frictionless inclined plane with a slope of 20 degrees.

connected to mA via frictionless, massless pulley and massless cord... mB = 3kg

http://img152.imageshack.us/img152/8286/imag0103l.jpg

Homework Equations

F= Mass * Acceleration

Acceleration = Mass / Force

The Attempt at a Solution

so far all I can get is the free body diagrams..
would this follow a = Fga + Fgb / ma + mb?

except I don't have force..

 

[azizlwl]

You have to draw 2 diagrams.
1. Forces exerted on mass#1
2. Forces exerted on mass#2

Make the unknown force as T.

 

[ksm2288]

I have drawn two diagrams? I attached a picture. So I'm just calculating the tension force?

 

[ehild]

Homework Statement

Block of mass mA = 2kg - lying on frictionless inclined plane with a slope of 20 degrees.

connected to mA via frictionless, massless pulley and massless cord... mB = 3kg

http://img152.imageshack.us/img152/8286/imag0103l.jpg

The normal force N is perpendicular to the slope. It is not vertical.
What do you call Fgma and Fgmb?
You drew only one force exerted to mb. You need to draw the force of gravity, too.
Remember that the tension, the magnitude of the force the cord exerts, is the same on both bodies.

Homework Equations

F= Mass * Acceleration

Acceleration = Mass / Force

The Attempt at a Solution

so far all I can get is the free body diagrams..
would this follow a = Fga + Fgb / ma + mb?

except I don't have force..

The equation is incorrect. You miss a few parentheses.
Explain, what Fga and Fgb are.


ehild

 

[Nickie]

Your attempt at a solution is on the right track, but there are a few things that need to be clarified. The equation F=ma is the equation for Newton's Second Law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, there are two masses and two forces to consider, so the equation would look like this:

Fnet = ma + mb

Where Fnet is the net force on the system (in this case, the two masses together), ma is the force on mass A, and mb is the force on mass B.

Now, let's look at the free body diagrams for each mass. For mass A, there are two forces acting on it: the force of gravity (mg) pulling it down the incline, and the normal force (N) perpendicular to the incline. Since the incline is frictionless, there is no force of friction to consider.

For mass B, there is only one force acting on it: the tension in the cord pulling it up.

Now, we can set up our equations:

For mass A:

Fnet = ma
ma = mgsin(20) - N
N = mgcos(20)

For mass B:

Fnet = mb
mb = T
T = mb

Now, we can substitute these equations into our original equation:

Fnet = ma + mb
Fnet = (mgsin(20) - N) + (mb)
Fnet = (mgsin(20) - mgcos(20)) + (mb)

We can then simplify this equation using the known values for mass A (2kg) and mass B (3kg):

Fnet = (2kg)(9.8m/s^2)(sin(20) - cos(20)) + (3kg)
Fnet = 2.73N + 3kg

So, the net force on the system is 5.73N. To find the acceleration, we can use the equation F=ma, where F is the net force and a is the acceleration:

F = ma
5.73N = (5kg)a
a = 5.73N / 5kg
a = 1.15m/s^2

So, the acceleration of the system is 1.15m/s^2.

In summary, the equation F=ma+mb is not quite correct