How Does Friction Affect the Motion of Blocks on an Inclined Plane?

[Exuro89]

Homework Statement

Two blocks with masses 4.00 kg and 8.00 kg are connected by a string and side down a 30.0 inclined plane. The coefficient of kinetic friction between the 4.00kg block and the plane is .25; tha between the 8.00kg block and the plane is .35.
(a) Calculate the acceleration of each block.
(b) Calculate the tension in the string.
(c) What happens if the positions of the blocks are reversed, so the 4.00kg block is above the 8.00 kg block?

[PLAIN]http://k.min.us/ilSZqW.png

Homework Equations

F=ma

The Attempt at a Solution

Okay so the first thing I do is do fbd of both masses. I then find the x and y component forces of each.

4kg block

F_y= F_n-mgcos(theta)=0
F_n=mgcos(theta)

F_x=T+F_k=ma
T+$$\mu$$_kmgcos(theta)-mgsin(theta)=-ma
T+$$\mu$$_kgcos(theta)-g(sin(theta)=-a

8kg block

F_y= F_n-mgcos(theta)=0
F_n=mgcos(theta)

F_x=F_k-mgsin(theta)-T=-ma
$$\mu$$_kmgcos(theta)-mgsin(theta)-T=-ma
$$\mu$$_kgcos(theta)-gsin(theta)-T=-a

Is this portion correct? I'm not sure how I can find each's acceleration with tension playing a part. If my math is right, without tension the first block would be going at 2.778m/s^2 and the larger block is going 1.9295m/s^2.

So with tension that would mean the first block is -a = T/4-2.778 and the second block -a = -T/8-1.9295. Is that all it's asking for for acceleration?

Would I then equal those equations to get T = 2.26N?

And for part C if the bocks were switched, T would be 0 and the 4kg block would run into the larger block. Is this correct?

 

[anathema]

Hello!

Your approach for finding the accelerations of each block is correct. However, there are a few minor mistakes in your calculations. I will walk you through the correct approach for each part of the problem.

(a) To find the acceleration of each block, we can use the equations you wrote for the x-component of the forces. However, we need to be careful with the signs. For the 4kg block, the equation should be:

T - μkmgcos(θ) - mgsin(θ) = ma

Note that the friction force should be subtracted since it acts in the opposite direction of motion. Also, the tension force should be positive since it acts in the direction of motion. Solving this equation for a, we get:

a = (T - μkmgcos(θ) - mgsin(θ))/m

For the 8kg block, the equation should be:

μkmgcos(θ) - mgsin(θ) - T = ma

Note that the friction force should be added since it acts in the same direction as the motion. Solving this equation for a, we get:

a = (μkmgcos(θ) - mgsin(θ) - T)/m

Now, we can plug in the values for μk, m, g, θ and T to find the accelerations of each block.

(b) To find the tension in the string, we can use the equations we derived for the accelerations. We can set the two equations equal to each other since the accelerations must be the same for both blocks (since they are connected by a string). This gives us:

(T - μkmgcos(θ) - mgsin(θ))/m = (μkmgcos(θ) - mgsin(θ) - T)/m

Solving for T, we get:

T = (μkmgcos(θ) - mgsin(θ))/2

Now, we can plug in the values for μk, m, g, and θ to find the tension in the string.

(c) If the positions of the blocks are reversed, the tension in the string will be 0. This is because the 8kg block will now be above the 4kg block, and there will be no tension in the string since the 8kg block will not be pulling on the string. The