Block sliding Down; find kinetic coefficient

  • #1

Homework Statement


A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 25.7°, the block starts to slide down the incline, traveling 3.50 m down the incline in 1.50 s. Calculate the coefficient of static friction between the block and the plank.
Motion of a block:
4.81×10-1

(THAT WAS THE FIRST PART OF THE QUESTION WHICH I HAVE ALREADY FOUND.)

Part 2:
Calculate the kinetic coefficent of friction between the block and the plank.


Homework Equations


None.


The Attempt at a Solution


not sure how but i keep getting: .21766

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2

Homework Statement


A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 25.7°, the block starts to slide down the incline, traveling 3.50 m down the incline in 1.50 s. Calculate the coefficient of static friction between the block and the plank.
Motion of a block:
4.81×10-1

(THAT WAS THE FIRST PART OF THE QUESTION WHICH I HAVE ALREADY FOUND.)

Part 2:
Calculate the kinetic coefficent of friction between the block and the plank.


Homework Equations


None.


The Attempt at a Solution


not sure how but i keep getting: .21766

Homework Statement





Homework Equations





The Attempt at a Solution

So, first, you have a FBD correct?

I'll assume that you do.

You should have a weight force going down.
A normal force perpendicular to the plank's surface.
And a friction force going the opposite direction of motion, correct?
 
  • #3
IDk what an FBD even is :(
 
  • #4
Google it real quick and you should get some good notes on it. :smile:
 
  • #5
Frozen Beverage Dispenser?!
just kidding. bhaha
Okay a Free Body Diagram. Well mmy professor told me to make one but I don't know how.
But okay, what you said makes sense. A normal force, a weight force and a friction force and then the motion.
 
  • #6
Right. Now, when you draw your free body diagram, you'll notice that N is slanted in a particular direction.

You can picture the x-y axis as being slightly rotated, that way the N is now directly on the y-axis, and the f force is now directly on the x-axis.

W is now a hypotenuse going down, with an angle between the W and y-axis of 25.7 degrees.

Before I go on, does the way I (tried) to explain the FBD make sense?
 
  • #7
Everything makes sense excpet where the W goes.
 
  • #8
Alright, so let me try it a different way.

Consider a normal x-y axis with N going upward directly on the y-axis and the f force going directly on the x-axis.

Now, your W force will be a diagonal line in the 4th quadrant (that is, negative y, positive x).

The angle between W and the y-axis will be 25.7 degrees. Does that make any more sense that way? :smile:
 
  • #9
That makes much more sense. :)
You're my new best friend.
 
  • #11
Haha. Thanks, just here to try to help.

Now, you need to find the sum of the forces in the x-direction.

Well, let's start by figuring out what N will be.

Since N is pointing straight upwards, and by Newton's 3rd law, N - Wcos(theta) = 0, right?

Now, N = Wcos(theta) -> W = mg, so...

N = mg*cos(theta)

f=[itex]\mu[/itex]N -> [itex]\mu[/itex]mg*cos(theta)....


[itex]\Sigma[/itex]F[itex]_{x}[/itex] = mg*sin(theta) - [itex]\mu[/itex]mg*cos(theta) = ma[itex]_{x}[/itex]

Correct?

If you see an error in my reasoning, correct me.
 
  • #12
Alright, most of that makes sense. i don't have the acceleration tho to find u?
 
  • #13
And all three masses cancel out right?
 
  • #14
Alright, well, that's what we need to do next. :smile:

And yes, all three masses cancel out.
 
  • #15
Okay. So now we have:
gSin(Theta) - u*g*cos(theta) = a
-->
4.25 - 8.83u = a
 
  • #16
Alright, most of that makes sense. i don't have the acceleration tho to find u?
Consider that the final velocity will be the Delta x / Delta t.

So, Vf = 3.50 m / 1.50 s

Vf = 2.333 m/s


Now, use your kinematic equation...

V[itex]_{f}[/itex]=V[itex]_{i}[/itex]+a*t

V[itex]_{i}[/itex]=0 m/s

Now, solve for a.....

a = [itex]\frac{V_{f}}{t}[/itex]

a = [itex]\frac{2.333}{1.50}[/itex]

a = ?


After that, you can solve for uk.
 
  • #17
Okay. So now we have:
gSin(Theta) - u*g*cos(theta) = a
-->
4.25 - 8.83u = a

That looks to be correct.
 
  • #18
it says i'm wrong....
i got a=1.553
so, u should equal .305
...right?
 
  • #19
That should be correct. (Unless we did something wrong in the procedure).
 
  • #20
we must have done something wrong...cuz it says i'm incorrect.
 
  • #21
g is always 9.8 ?
and theta is always 25.7 ?
 
  • #22
g is always 9.8 ?
and theta is always 25.7 ?

g is approximately 9.8 m/s^2

Theta will be 25.7 degrees in this problem.

uk may be .322.... or .323. I'm not quite sure right now.
 
  • #23
Then i have no idea what to do. i relooked my math, and it makes sense to me.
 
  • #24
Then i have no idea what to do. i relooked my math, and it makes sense to me.

Maybe you could multiply the 1.553 m/s^2 by cos(25.7). Since a would be a vector, and that would, technically, give you ax.

Try that and see if that works for you. I'm honestly confused.
 

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