Dynamics ramp and friction. Finding angle?

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Homework Help Overview

The problem involves a 3.0 kg block sliding down a rough ramp with a specified acceleration and a known force of kinetic friction. The objective is to determine the angle of the ramp based on the given parameters.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the role of friction in the context of the problem. There are questions about the correct substitution of values for the frictional force and the setup of the equations. Some participants note potential sign mistakes in the equations used to relate forces.

Discussion Status

The discussion is ongoing, with participants providing insights into the equations and the need for careful consideration of force directions. There is recognition of the complexity introduced by the frictional force and the acceleration, and some guidance has been offered regarding the correct formulation of the equations.

Contextual Notes

Participants are navigating the implications of the given values, such as the force of friction and the acceleration, while also addressing the need for clarity in the sign conventions used in their calculations.

britt6

Homework Statement


A 3.0 kg block is released from the top of a rough ramp of length 2.0m and accelerates down the ramp at 1.6m/s^2. If a force of kinetic friction of 10 N acts on the block, what is the angle of the ramp.

Homework Equations


Fnet=ma
ff=(mu? I don't have that symbol HAHA) muFn

The Attempt at a Solution


m=3.0kg
l=2.0m
a=1.6m/s^2
ff=10Nif Fnet=ma... then

ff+fgparallel=ma
ma-ff=fgparallel
ma-ff=mgsintheta
ma-ff/mg=sintheta
then take inverse to get theta..

Now what! answer should be 30 but when I put in the values I get 10 degrees. help please
 
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britt6 said:
ma-ff/mg=sintheta
What are you plugging in for ff here? This may be where the mistake is.
 
NFuller said:
What are you plugging in for ff here? This may be where the mistake is.

I just thought because the question states that there is a kinetic force of friction and acceleration, there are unbalanced forces, meaning you need an Fnet. Therefore I plug in Ff kinetic to complete that idea.
 
britt6 said:
Therefore I plug in Ff kinetic to complete that idea.
Oh, right. I missed that they gave you the force of friction in the problem.
britt6 said:
ff+fgparallel=ma
You have a sign mistake here, it should be
$$-F_{f}+F_{g}=ma$$
because the vectors for friction and gravity point in opposite directions.
 
britt6 said:
mu?
When editing a post, there is a line of symbols above the text area. If you click on the Σ symbol a couple of lines of special characters will appear below the text area. Click those as necessary.
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NFuller said:
You have a sign mistake here, it should be
$$-F_{f}+F_{g}=ma$$
because the vectors for friction and gravity point in opposite directions.
That depends on the sign convention chosen. If Britt chooses the same direction as positive for all forces and accelerations (a good approach often) then the equation for parallel to the plane is $$\Sigma F=F_{f}+F_{g}=ma$$
The negative sign comes in when values are plugged in. If downslope is positive then Ff=-10N, Fg=+mg sin(θ).
 

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