Motion on Inclined Plain with Pulley

[magixx]

Homework Statement

Cart is on an inclined plain, attached to the cart is a weightless string going to a pulley attached to another mass. There is no friction.
Cart Weight: 0.5Kg
Mass Weight: 0.1Kg
Plain Angle: 20 Degrees
Diagram: See below

Question: Determine Tension in string and acceleration.

Homework Equations

Fg=ma

The Attempt at a Solution

http://img254.imageshack.us/img254/6286/physji3.jpg

With the two equations I then made them both equal 5a and then used elimination, I got T=0.6582N and a=3.21m/s.
I kinda doubt this is the right answer.

 

[LowlyPion]

Homework Statement

Cart is on an inclined plain, attached to the cart is a weightless string going to a pulley attached to another mass. There is no friction.
Cart Weight: 0.5Kg
Mass Weight: 0.1Kg
Plain Angle: 20 Degrees
Diagram: See below

Question: Determine Tension in string and acceleration.

Homework Equations

Fg=ma

With the two equations I then made them both equal 5a and then used elimination, I got T=0.6582N and a=3.21m/s.
I kinda doubt this is the right answer.

Welcome to PF.

I think you need to be consistent in how you assign signs between the two equations.

The force on your weight should be more properly given by .1a = T - .98

 

[baba89]

I would approach this problem by first analyzing the forces acting on the system. In this case, we have the force of gravity (Fg) acting on both the cart and the mass, and the tension force (T) acting on the string. Since there is no friction, there are no other forces to consider.

Using the equation Fg=ma, we can calculate the force of gravity acting on each object. For the cart, Fg=0.5kg x 9.8m/s^2 = 4.9N. For the mass, Fg=0.1kg x 9.8m/s^2 = 0.98N.

Next, we need to consider the direction of these forces. The force of gravity acts straight down for both objects, while the tension force acts in the direction of the string. Since the string is attached to the cart and the mass, the tension force will be pulling in opposite directions for each object.

For the cart, the tension force will be pulling up the incline, at an angle of 20 degrees. We can break this force into its horizontal and vertical components by using trigonometry. The horizontal component of the tension force will be Tcos20, and the vertical component will be Tsin20.

For the mass, the tension force will be pulling down the incline, at an angle of 20 degrees. Again, we can break this force into its horizontal and vertical components, which will be -Tcos20 and Tsin20 (since the force is acting in the opposite direction for the mass).

Now, we can set up two equations using Newton's second law (F=ma) for each object.

For the cart: Tcos20 = 4.9a
For the mass: Tsin20 - 0.98 = 0.1a

We have two equations and two unknowns (T and a), so we can solve for both. First, we can solve for T by isolating it in one of the equations. Let's use the second equation: Tsin20 = 0.1a + 0.98. Now, we can plug this into the first equation to solve for a: (0.1a + 0.98)cos20 = 4.9a. Solving for a, we get a= 3.21m/s^2.

To find the tension force