A block of mass 2 kg rests on an inclined plane of angle 30 degree. The coefficent of static friction between?

Question

block and incline is 0.35. The block is attached by a cord that lies over a pulley at the top of the incline to another hanging block. if the system is to remain at rest, what is the maximum mass of the hanging block?

Answer 1

To answer this question, you need to draw a free body diagram. Unfortunately, it's impossible to draw one on this website. If you'd like to see the one I came up with, feel free to IM me!


We need to break all of the forces into x and y components.
The 2 kg block has the following forces acting on it.

T1 = Tension in the string
fr = Friction
N = Normal force
mg = Weight

The hanging mass has the following forces
T2 = Tension in the string
Mg = weight


One trick we can use to make our calculations a bit simpler is to look at each block with a different set of axis.


For the 2 kg block, let the x axis run parallel to the incline. Then we have the following equations (notice I haven't plugged in any values).

y - direction
N = mg * sin(30º)

x - direction
mg * cos(30º) = T1 + fr


For the hanging block, let the axis remain vertical like we're used to.

y - direction
Mg = T2

x - direction
None!


Ok, we're almost done. First, T1 = T2 since we're assuming a massless string with no slack. Second, We know that

M = T / g, so we need to find the tension.

The first set of equations will come to the rescue!

Since we have friction, we need to know the normal force, so let's calculate that.


N = mg * sin(30º) = 2 * 9.8 * sin(30º) = 2 N

Now we can use the second equation to determine the tension in the rope.

mg * cos(30º) = T + fr

2 * 9.8 * cos(30º) = T + 0.35 * 2

T = 16.274 N

Finally, using the fact

M = T / g, we get

M = 16.274 / 9.8 = 1.66 kg!

There ya go :D