Inclined plane and normal reaction force

[robvba]

A block of mass M slides down a frictionless plane inclined at an angle è with the horizontal. The normal reaction force exerted by the plane on the block is

a Mg.
b Mg sin è.
c Mg cos è.
d zero, since the plane is frictionless

Homework Equations

Fn=mg cos è

The Attempt at a Solution

c

is this right?

thank you

 

[Redbelly98]

Here is a simple way to check:

What is the normal force when the angle is zero (horizontal surface)?
What is the normal force when the angle is 90° (vertical surface)?

Which choice is consistent with those two cases?

 

[robvba]

Here is a simple way to check:

What is the normal force when the angle is zero (horizontal surface)?
cos (0) = 9.8
What is the normal force when the angle is 90° (vertical surface)?
sin (90) = 9.8
Which choice is consistent with those two cases?

9.8 is consistent. However, the book says, "unless angle = 0, Fn has magnitude less than the weight mg."

 

[Redbelly98]

I don't understand your response.

Are you sure the normal force is 9.8 when the face is vertical?

And what do you mean "cos (0) = 9.8"? The cosine of zero is 1.

However, the book says, "unless angle = 0, Fn has magnitude less than the weight mg."

Your book is correct.

Edit:
You were also correct in your 1st post.

 

[robvba]

Here is a simple way to check:

What is the normal force when the angle is zero (horizontal surface)?
What is the normal force when the angle is 90° (vertical surface)?

Which choice is consistent with those two cases?

I don't understand your response.

Are you sure the normal force is 9.8 when the face is vertical?
heh, i must have been tired and confused. i was multiplying 9.8 by cos and sin.
And what do you mean "cos (0) = 9.8"? The cosine of zero is 1.


Your book is correct.

Edit:
You were also correct in your 1st post.

thank you