Newton's Forces - Block Down An Incline

AI Thread Summary
The discussion focuses on solving a physics problem involving a block sliding down an incline at a 25.5° angle with a coefficient of kinetic friction of 0.17. The key steps include breaking down the forces into components, recognizing that the mass cancels out in the equations, and applying Newton's second law to find the acceleration and final speed of the block. The correct acceleration was determined to be 2.715 m/s², and the final speed at the bottom of the incline was calculated to be 7.18 m/s. The importance of accurately drawing free body diagrams to visualize forces was emphasized. The conversation also briefly shifted to a new topic regarding forces acting on blocks on a frictionless surface.
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[a]1. Homework Statement [/b]

The block shown in Fig. 5-27 lies on a smooth plane tilted at an angle = 25.5° to the horizontal, with µk = 0.17.

5-27.gif


Figure 5-27

(a) Determine the acceleration of the block as it slides down the plane.
________ m/s2 (down the plane)
(b) If the block starts from rest 9.50 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?
_________ m/s

Homework Equations



F=m(a)
Ff=Mk(N)

[c]3. The Attempt at a Solution [/b]

I drew a Vector Diagram so that gravity was pointing directly downward from the box, so that means that the Fg vector is at an angle of -64.5 from the Horizontal, (or 25.5 degrees from the Y axis, whichever you prefer!), and then so Fg has horizontal and Vertical components.

Fy(net)=m(a)=Fa+Fgx-Ff
Ff=.17(Fnormal)

? I'm very confused..it seems like I need to know the mass to solve the problem, but the other half of me doesn't even know where to begin.
 
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First, you need to break down the forces into components just as you said.
forces in the Y direction will be mgcos(theta)-Normal Force=0 since the box is not rising or sinking.

This tells you that mgcos(theta) is therefore equal to normal force.

Now forces in the X direction will be mgsin(theta)-Kinetic Friction=ma (it equals ma since it is moving)

We know that kinetic friction = Normal force*friction coefficient (by definition). We also know from above that normal force is equal to mgcos(theta).

So continuing to solve forces in the X direction - you will have mgsin(theta)-force of kintetic friction=ma. Here we will use mgcos(theta) to stand in for normal force.

mgsin(theta)-mgcos(theta)(0.17)=ma

now notice how the masses will cancel since they are on both sides.

gsin(theta)-gcos(theta)(0.17)=a

edit: I think your confusion might come from a weakness in drawing clear free body diagrams. I will try and sketch one for you in a paint program.
 
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Hope this isn't too confusing. I am not a graphic artist.

http://img141.imageshack.us/img141/686/fbdue2.gif

The key point is that when you draw the mgsin(theta) force - make sure it is parallel to the slope. Some beginners draw it as a flat 90 degree line and they can't visualize the forces correctly.
 
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Hey! Wow, you helped me so much! I got the answer right on the first try! =) (the acceleration is 2.715 by the way).

I realized what I did wrong, by the way. I didn't realize that the forces in the Y direction equaled zero, although I should have. And I was discouraged by the fact that we didn't know mass, but it ended up canceling out at the end. Thanks again! =)

I even got the 2nd part too: it was 7.18 by the way. ;)

Thanks again Bob! You were great!
 
Glad I could help.
 
I want to know about the Forces acting on each other; for 3 blocks sitting side by side
on a frictionless leveled table top. 2 blocks on the sides are of same mass. The one on the middle is bigger than the tow on the sides.
 
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