# Is the Friction Force Affected by an Inclined Force?

• Alex126
In summary: No, it is not simply a weight component on an incline plane. In summary, the body will have a velocity of 1.2 m/s2 after a period of time t has elapsed.
Alex126

## Homework Statement

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

F = m*a

## 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?

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

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.

Something like this...

#### Attachments

• Block.PNG
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Alex126 said:
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)]
Neither. What equations have you been taught regarding frictional forces? (Should be listed under relevant equations.)

Alex126 said:
Force F_X = Force F * cos (α)
Yes.
Alex126 said:
Weight force pointing down, and a Normal force pointing up; I omitted them, as they shouldn't matter.
They matter.
Alex126 said:
do the forces on the Y axis really = 0 (= no motion) ?
Yes.
Alex126 said:
the Normal force of the plane is actually lower by -Force F once the Force F is applied.
Not quite. Stick to vertical components.

Alex126 said:
velocity v = 1.2 m/s2
I assume you mean m/s.

haruspex said:
I assume you mean m/s.
Yea, that was it, my bad >_<

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

haruspex said:
They matter.
#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?

haruspex said:
Not quite. Stick to vertical components.
Oh, I see, I see. Let me try this again :D
the Normal force of the plane is actually lower by -Force F_Y once the Force F is applied.

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

Hello?

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

Ok, that's better. Thanks.

## 1. What is a block with "incline force"?

A block with "incline force" is a type of block that is placed on an incline or slope. It experiences a force due to gravity, known as the incline force, which is the component of the force of gravity that acts parallel to the surface of the incline.

## 2. How does the incline force affect the block?

The incline force affects the block by causing it to accelerate down the incline. This is because the incline force is acting in the same direction as the slope, increasing the overall force on the block and causing it to move.

## 3. What factors affect the magnitude of the incline force?

The magnitude of the incline force is affected by the mass of the block, the angle of the incline, and the acceleration due to gravity. The greater the mass of the block or the steeper the incline, the larger the incline force will be. The acceleration due to gravity also plays a role, as it determines the strength of the force of gravity acting on the block.

## 4. How is the incline force calculated?

The incline force can be calculated using the formula F = mgsinθ, where F is the incline force, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline. This formula takes into account the factors that affect the magnitude of the incline force.

## 5. What are some real-life applications of the incline force?

The incline force is commonly seen in everyday situations, such as when pushing a cart up a ramp or when sliding down a slide. It is also important in engineering, as it is a crucial factor in designing structures that can withstand forces on inclined surfaces, such as bridges or ramps. Additionally, understanding the incline force can help in understanding the motion of objects on slopes and predicting their behavior.

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