# Block with "incline Force" - 3

1. Jan 28, 2017

### Alex126

1. The problem statement, all variables and given/known data
A block of mass m1 (8.5 kg) lies on a flat (non-incline) plane, with friction coefficient μ (0.2). It is pulled by a force F (32 N) that has makes an angle α (22°) with the plane. Determine:
- Acceleration
- After how much time t the body will have a velocity v = 1.2 m/s2

2. Relevant equations
F = m*a

3. The attempt at a solution
The doubt I have is in regards to what component of the force F we should consider. Here's a drawing:

I know there is also a Weight force pointing down, and a Normal force pointing up; I omitted them, as they shouldn't matter.

First, I need to write the forces on the body, and separate those on the X axis from those on the Y axis. I'll choose +Y going up, and +X going right.

X. [Friction] + [Force F_X] = m*a
Y. [Weight] + [Normal] + [Force F_Y] = 0 (since there is no vertical movement)

So here's the two stupid questions that I have.

Q1: Friction force = [coefficient] * [weight force_Y] ? I would have to say so, but for some reason in my mind now I have a doubt that it might be:
Friction force = [coefficient] * [weight force] * [cos (α)]

In other words, I don't know if I should take into consideration the fact that the force is "inclined" or not. My guess would be not...but my brain refuses to cooperate here, so please just give me the correct answer since I'll never get it on my own.

So, it's either A or B:
A. Friction force = 0.2*8.5*9.81 = 16.7
B. Friction force = 0.2*8.5*9.81* cos (22°) = 15.5

Q2: assuming we know the correct Friction force, now we need to find the acceleration. Obviously I'll be using the equation on the X axis, but I need Force F_X, which is the X-axis component of Force F. Correct?

If so, then I would say that Force F_X = Force F * cos (α)
That's because of this triangle here:

This sounds obvious and straight-forward to me, but I need confirmation please.

So at the end it would be:
a = (-Friction + Force F_X) / (m)
a = (-16.7 + 32*cos (22°) ) / (8.5)
OR
a = (-15.5 + 32*cos (22°) ) / (8.5)

Q3: do the forces on the Y axis really = 0 (= no motion) ? I would assume so, but here's the thing. If without Force F we have that Weight + Normal = 0, then why is it that with +Force F we still have = 0? If I had to guess I would say that it's because one between Weight and Normal goes down by -Force F, and since Weight is fixed, then the Normal force of the plane is actually lower by -Force F once the Force F is applied. Is this correct?

2. Jan 28, 2017

### Arman777

Could you use latex or at least sembols.Its really hard to read

3. Jan 28, 2017

### Vector1962

May I suggest you draw a sketch that labels all the forces? The sketches you have made are close, but are missing at least one force. It is very important to start with a nice sketch that labels ALL the forces.

4. Jan 28, 2017

### Vector1962

Something like this...

#### Attached Files:

• ###### Block.PNG
File size:
4.3 KB
Views:
20
5. Jan 28, 2017

### haruspex

Neither. What equations have you been taught regarding frictional forces? (Should be listed under relevant equations.)

Yes.
They matter.
Yes.
Not quite. Stick to vertical components.

6. Jan 28, 2017

### haruspex

I assume you mean m/s.

7. Jan 29, 2017

### Alex126

Yea, that was it, my bad >_<

We've been taught that friction = [weight component on the Y axis]*coefficient, but that was in slightly different problems where there was a block on an incline plane.

It might have also been friction = [Normal force]*coefficient, which I'm starting to believe might be the case due to the next quote.

#Yforcesmatter

I thought they didn't matter because the Weight force was nullified by the Normal force. Can you elaborate as to why they matter?

Oh, I see, I see. Let me try this again :D
So, since you mentioned that neither [Weight]*coefficient, nor [Weight]*coefficient*cos (22°) is correct, my next guess (I'm trying to read between the lines since nobody gave the direct answer) would be that friction force = normal*coefficient. So...(drumroll)

Friction force = ( [Weight]-[Force F_Y] ) * coefficient
(Where [Weight]-[Force F_Y] = "new" Normal force after Force F is applied)

Better, worse, or just as wrong? :D

8. Jan 29, 2017

### Alex126

Hello?

9. Jan 30, 2017

### Vector1962

You said it right, the friction force Ff = μN. You find N by ∑Fy (generally speaking).

10. Jan 30, 2017

### Alex126

Ok, that's better. Thanks.