How Do You Solve Inclined Plane Motion Problems?

[nysnacc]

Homework Statement

Homework Equations

f = mu_k * N
F=ma

The Attempt at a Solution

 

[phinds]

This is the second post you have made with just your working out a problem but no question. The natural assumption is that you are asking for verification of your result, BUT ... you could be after something more specific, so it would be a good idea to ask whatever it is that you actually want to know.

 

[nysnacc]

Yes, I have something to ask.
First, the y-component of F is part of the force contribute to the normal force, right?
I want to know if I need to consider the friction when the box is pushed up to the inclined plane, which gives: F cos 20 - mg sin 20 -f...
then when the box is sliding downward, there's another friction against the motion... then the sum of force in x-dir will be:
[F cos 20 - mg sin 20 -f] +f = F cos 20 - mg sin 20

Am I correct? so then, I use this divide by m to get a, and integrate to get v

Thanks a lot

 

[gneill]

It looks like your method for finding the net force acting along the slope is okay. But you may want to recalculate the friction force; The result doesn't match what I think it should be. Perhaps a calculator slip? I see that you've made one of the angles 30° in your work, but that's probably just a typo, right?

Once you've found the acceleration you shouldn't need to integrate if you know your basic kinematic formulas.

 

[nysnacc]

It looks like your method for finding the net force acting along the slope is okay. But you may want to recalculate the friction force; The result doesn't match what I think it should be. Perhaps a calculator slip? I see that you've made one of the angles 30° in your work, but that's probably just a typo, right?

Once you've found the acceleration you shouldn't need to integrate if you know your basic kinematic formulas.

Oh hey, I made an mistake, as you said, wrote 30° instead of 20°... and the answer for f is actually wrong.

What is the formula for the sum of force in x -direction (my x coordinate is not horizontal) ??
Cuz, I need to find v, what will be a better way if not integrating acceleration?

Thanks a lot

 

[gneill]

What is the formula for the sum of force in x -direction (my x coordinate is not horizontal) ??

You've been finding the components of the various forces along the x and y directions using the trig functions. So you already have them in hand. The friction force is directed along the slope, for example.

Cuz, I need to find v, what will be a better way if not integrating acceleration?

If you have the net force acting on mass M along the direction of the slope, what's the acceleration of M in that direction? Knowing acceleration the rest is basic kinematics.

 

[nysnacc]

It is force applied - weight - friction? but the mass's motion is actually opposite of the direction of force applied

 

[nysnacc]

I mean: net force = force applied - weight - friction,
then divide m to get a

 

[gneill]

I mean: net force = force applied - weight - friction,
then divide m to get a

That's right, assuming you mean the components of F and g force that are directed along the slope. It's the net force that causes the mass to accelerate. Note that it's entirely possible that F will be insufficiently large to prevent the mass from sliding downslope rather that upslope! You should check the magnitudes of the upslope F component against the downslope g component prior to "adding" you friction. Friction always opposes the direction of motion.

 

[nysnacc]

That's right, assuming you mean the components of F and g force that are directed along the slope. It's the net force that causes the mass to accelerate. Note that it's entirely possible that F will be insufficiently large to prevent the mass from sliding downslope rather that upslope! You should check the magnitudes of the upslope F component against the downslope g component prior to "adding" you friction. Friction always opposes the direction of motion.

Yes, I meant the net force in x dir.

As we push the box, up the slope, there is a friction opposing, but the force applied is not strong enough, so the box slide down. Is there a friction (in -x direction) opposing the motion?

 

[gneill]

Yes, I meant the net force in x dir.

As we push the box, up the slope, there is a friction opposing, but the force applied is not strong enough, so the box slide down. Is there a friction (in -x direction) opposing the motion?

If the box slides downslope then the friction will oppose that motion. There's only ever one friction force operating between surfaces and it always opposes the motion that is taking place (or, in the case of static friction, trying to take place).

If the motion is downslope, the friction will oppose this and be directed upslope.

 

[nysnacc]

So the true net force equation will then be: F_x = Fcos20 - mgsin20 +f

 

[gneill]

That looks good.

 

[nysnacc]

That looks good.

Then I can divide mass to get a..
and then integrate a = get v?

 

[nysnacc]

which dv/dt = F/m

∫ dv/dt *dt= ∫ F/m dt

v-v₀ = a*t

 

[nysnacc]

which v₀ = 0 (at rest)

so v(t=2) = a*2 m/s

 

[gneill]

Then I can divide mass to get a..
and then integrate a = get v?

Sure, but it's easier to apply standard kinematics equations. Remember SUVAT?

 

[nysnacc]

Sure, but it's easier to apply standard kinematics equations. Remember SUVAT?

Nope, Like a= v_r +v_θ thingy?

 

[nysnacc]

OHH v=u+at

 

[gneill]

Like v = v_o + at. You know the acceleration, you know the time. The initial velocity is zero. So...

 

[nysnacc]

I got it!

 

[nysnacc]

Thanks!

 

[gneill]

OHH v=u+at

Ah. The flash of insight!