How Is Tension Calculated in a Two-Object System on an Inclined Plane?

[phish]

Homework Statement

An object on an inclined ramp of mass 4 kg forms an angle of 34° with the horizontal. The object on the ramp is connected to a second object of mass 6.4 kg on a horizontal surface below an overhang that is formed by the inclined surface. An external force of 30 N is exerted on the object on the ramp. Both objects are accelerating. Assume that the surfaces and the pulley are frictionless, and the connecting string and the pulley are massless, what is the tension in the string connecting the two objects?

Homework Equations

F = ma
m1g - T = m1a
T - m2g = m2a

The Attempt at a Solution

I drew out the FBD for the object on the incline

The force of gravity of the object must be broken into component parts:

(4)(9.8) = 39.2 N
x component: 39.2 sin34 = 21.9 N
y component: 39.2 cos34

The x component is counteracting the exerted force of 30 N, so 30-21.9 = 8.1 N

object 2 is not on an incline so: (6.4)(9.8) = 62.7 N is the force applied on it

8.1 - T = 4a
T - 62.7 = 6.4a

From here I am lost because if the above equations are solved for a - it turns out to be negative.

Any help in finding acceleration and then tension would be greatly appreciated
Thanks

 

[phish]

diagram

 

[tomcue]

for providing the problem and your attempt at a solution. I would approach this problem by first identifying the key variables and equations involved. From the given information, we have the masses of the two objects (4 kg and 6.4 kg), the angle of the inclined ramp (34°), the external force acting on the object on the ramp (30 N), and the fact that all surfaces and the pulley are frictionless and massless.

Based on these variables, we can use Newton's second law (F=ma) to determine the acceleration of the two objects. However, we need to consider the forces acting on each object separately. For the object on the incline, the forces acting on it are its weight (mg) and the tension in the string (T). The weight can be broken down into its x and y components, as you have done in your attempt. The y component of the weight is balanced by the normal force from the ramp, so the net force in the y direction is 0. This means that the acceleration in the y direction is also 0.

In the x direction, we have the external force (30 N) and the x component of the weight (21.9 N) acting in the positive direction, and the tension (T) acting in the negative direction. So we can set up the following equation:

30 N + 21.9 N - T = (4 kg)a

For the object on the horizontal surface, the only force acting on it is its weight (62.7 N). So we can set up the following equation:

T - 62.7 N = (6.4 kg)a

Now we have two equations with two unknowns (T and a). We can solve for T by substituting the second equation into the first:

30 N + 21.9 N - (6.4 kg)a - 62.7 N = (4 kg)a

-10.8 N - (6.4 kg)a = (4 kg)a

-10.8 N = (10.4 kg)a

a = -1.04 m/s^2

Now we can substitute this value of a into the second equation to solve for T:

T - 62.7 N = (6.4 kg)(-1.04 m/s^2)

T = 55.7 N

So the tension in the string connecting the two objects is