Block on Ramp HW: Normal Force & Static Friction

[012983]

Homework Statement

I did one of the standard block-on-a-ramp labs last week. It involved placing a pre-defined mass on a plank of wood and raising the plank until the mass started to slide down. I recorded theta and rolled from there.

I'm kind of looking for a confirmation of my work thus far and a little help on how to go on. I have the mass down to the x- and y-components and have to determine the normal force, and from there, force and the coefficient of static friction.

The table I'm using is attached.

Homework Equations

F = ma
F(s) = U(s)*N

The Attempt at a Solution

I used mgcos(theta) for the y-component and mgsin(theta) for the x-component of the weight.

I'm thinking the normal force is mgcos(theta).

I assume I use F=mgsin(theta)=ma to solve for the next column in the table. Do I get a from a = gsin(theta) (acceleration along the incline)?

I understand how to get the static force in the last column. Thanks for any help.

 

[amy andrews]

You've calculated the x and y components of the weight correctly; mg_x=mg sin(theta) and mg_y=mg cos(theta).
The normal force is also correct; F_n=mg cos (theta).
However, you've forgotten that ma is equal to the sum of all forces in a system. Therefore (you must take the forces in the x direction only, since the forces in the y direction simply sum to zero):
Sum of forces_x= ma
mg_x-F_fr=ma
The equation for friction force, where u is the coefficient of friction, is u*F_n. In this case:
mg_x-u*F_n=ma
mg sin(theta)-u*mg cos (theta)=ma
Cancel the masses, factor out the g:
a= g*[sin (theta) - u cos (theta)].
Hope this helps!

 

[012983]

You've calculated the x and y components of the weight correctly; mg_x=mg sin(theta) and mg_y=mg cos(theta).
The normal force is also correct; F_n=mg cos (theta).
However, you've forgotten that ma is equal to the sum of all forces in a system. Therefore (you must take the forces in the x direction only, since the forces in the y direction simply sum to zero):
Sum of forces_x= ma
mg_x-F_fr=ma
The equation for friction force, where u is the coefficient of friction, is u*F_n. In this case:
mg_x-u*F_n=ma
mg sin(theta)-u*mg cos (theta)=ma
Cancel the masses, factor out the g:
a= g*[sin (theta) - u cos (theta)].
Hope this helps!

I thought of using that equation, but I didn't have a value for u to plug in... I assumed we were supposed to find a another way and use it to find u.

Now I'm confused even more, heh. Thanks for the response, though!

 

[amy andrews]

You can find u from
mg_x>F_fr, since the block will only start sliding when the x component of the weight is greater than the force friction.
Does that help at all?

 

[012983]

You can find u from
mg_x>F_fr, since the block will only start sliding when the x component of the weight is greater than the force friction.
Does that help at all?

I recalculated the values for u and got the exact same values as I did using a = gsin(theta). However, when I recalculate the net force I get very small results (e.g., 0.0000847 N) and in one case a negative net force. Is this right?

Thanks.

 

[amy andrews]

I'm not completely sure what you mean by "net force". In the y direction, however, the forces should sum to zero, as the force normal cancels out the y component of the weight. Your extremely small force should possibly be simply zero- perhaps you rounded off too soon when calculating the forces?
For the negative force- remember that positive and negative forces simply depend on direction. I'm assuming you're taking "down the ramp" as negative and "up the ramp" as positive. In that case, when the x component of the weight becomes larger than the force friction, the net force will indeed be negative.
I'm not sure if this was what you were looking for...let me know!