Monkey and sled - conservation of energy problem

[physicsstudent06]

d = unknown

u = 0.20

 

[physicsstudent06]

i still don't know how to find the normal force, or the force due to friction

 

[Doc Al]

To find the normal force, realize that it must exactly balance the component of the weight that is perpendicular to the incline. (The weight and normal force are the only two forces on the monkey/sled that have components perpendicular to the incline.)

 

[physicsstudent06]

weight force = 196 N

 

[physicsstudent06]

so is the normal force equal to the weight force?

 

[Doc Al]

so is the normal force equal to the weight force?

No. Re-read my previous post. (Pay attention to the word "component".)

Try this: A 1kg book lays on a horizontal table. What's the normal force? In this case, the normal force does equal the weight.

But what if the table is tilted? What happens to the normal force then?

 

[physicsstudent06]

ok, so the normal force would be 1 kg for that book.

in our scenario, normal force would equal weight times sin theta?

 

[physicsstudent06]

or weight plus sin theta

 

[physicsstudent06]

i don't know, I'm frustrated

 

[physicsstudent06]

is the normal force mass x gravity x sin (theta)?

 

[physicsstudent06]

sin(theta) would be the height over the hypotenuse. the hypotenuse is d, but we don't know the height

 

[physicsstudent06]

gotta go, but thanks for your help so far

 

[Doc Al]

is the normal force mass x gravity x sin (theta)?

Here's another hint. The parallel & perpendicular (to the incline) components of the weight are:
W_\parallel = -mg \sin \theta
W_\perp = - mg \cos \theta

 

[Hootenanny]

Firstly let my clarify my comment which seems to be causing som confusion;

Careful, the normal force always act perpendicular to the surface. Gravity in this case is not acting perpendicular to the surface.

I said, that gravity is not acting perpendicular to the plane, I never said that it is not acting at all! As Doc Al said, there will be a component(part) of gravity which is acting parallel to the plain and one which is acting perpendicular to the plane, you must take into account these forces. As Doc Al and I have said, I think that it would be useful to draw a diagram. Now, I have found a diagram that illustrates the compoents pretty well (I haven't got time to draw one myself). Note that the vin the image is not a force, it is simply illustrating the direction of the velocity.
http://img149.imageshack.us/img149/6381/inclinedplane8xd.jpg
Taken from the PIRA website

So let us now take stock. There are X components of forces acting;
(1)Friction - parallel to and down the inclined plane.
(2)Normal Reaction force - perpendicular to the plane.

And we have two components of gravity; parallel (W_\parallel) and perpendicular (W_\perp), the equations for which Doc Al has supplied above. Now the component of weight acting parallel to the plane is acting downwards, in the opposite direction of the velocity and the same direction as the frictional force. The component of weight acting perpendicular to the surface is acting in the opposite direction to the normal reaction force (N in the above diagram). Now, as I said in a previous post, as the block is only accelerating in the plane parallel to the incline, this implies that the sum forces perpendicular to the plane must be zero. This implies that N = W_\perp. Do you follow? Can you go from here?