Homework Help: Incline and Pulley - 2

1. Jan 28, 2017

Alex126

1. The problem statement, all variables and given/known data
On an incline plane of known angle α (30°) lies a block of mass m1 (0.23 grams), connected through a pulley without friction to a second block of mass m2 (0.18 grams). Determine:
- The acceleration of the block
- The Tension force
- The Normal force of the plane

2. Relevant equations
F = m*a

3. The attempt at a solution
The simplest of the requests is actually the one that troubles me the most. The Normal force of the plane should be:
Normal = -Weight1_Y
(Weight1_Y = Y-axis component of the weight force of the first block)
Therefore, Normal = Weight 1 * cos (30°), so Normal = 0.23*9.81*0.87 = 1.96

As simple as it should be, I have some doubts that there might be more to it, such as some sort of influence on the Normal force by the second object in the system.
So, in other words, I'm asking if the Normal force in this case is still just -Weight1_Y, or if there's something else to consider.

Also, note that the problem asks the "Normal force of the plane", so if there is any other "Normal force", other than the plane's reaction to the Weight force (Y component only) of the body that lies on it, please let me know.

First I have to write the forces for each body individually.

1. [Tension] + [Weight 1] + [Normal] = m1*a
2. [Tension] + [Weight 2] = m2*a

Choosing +X towards the motion (assuming block m1 is sliding down), and +Y upwards (assuming block m2 is going up), this would become:

1. -T + W1_X + W1_Y - Normal = m1*a
2. +T - W2 = m2*a

The #1 becomes (W1_Y - Normal = 0): T = W1_X - m1*a
The #2 becomes: T = W2 + m2*a
Solving the system, we have W1_X - m1*a = W2 + m2*a, so:
a = (W1_X-W2) / (m1+m2)
a = (0.23*9.81*sin (30°) - 0.18*9.81) / (0.23-0.18)
a = -1.55

Since acceleration is negative, it means that the motion happens in the direction opposite the chosen axis system. Therefore, I conclude that the block m1 doesn't actually slide down the incline plane, but actually goes up the plane with an acceleration of +1.55 m/s2

Normally I wouldn't have a problem with this conclusion, but I just want to make sure.

Anyhow, the Tension is then calculated from the same system, for instance the first equation, so:
T = 0.23*9.81*sin (30°) - 0.23*(-1.55)
T = +1.48
Same number results from T = W2+m2*a (0.18*9.81+0.18*(-1.55))

2. Jan 28, 2017

goodphy

3. Jan 28, 2017

haruspex

Fundamental principle: each rigid body in the system responds to those forces which act on it directly. It cannot know or care about other forces. This is why it's a good idea to draw a separate free body diagram for each rigid body in the system.
Not that I can see.
Right.

4. Jan 29, 2017

Ok, thanks.