Coulomb's Law Application: A charge repelling a mass on a frictionless incline

[DanielGuh]

Homework Statement

At what distance from the 4 x 10-6 C charge is the 2.30 gram mass shown below (file attached) in equilibrium?

The angle of the frictionless incline is 28 degrees.

a) 1.31cm
b) 4.51m
c) 1.60m
d) 0.451m

Relevant Equations

Fg=mgcos(theta)
Fc=kq1q2/r^2
electrostatic constant (k) = 8.99*10^9

Fg = Fc
Fg = 2.3g*9.8m/s^2*cos(28)=19.90N
19.90 = (8.99*10^9)*(4*10^-6)*(6*10^-6)/(r^2)
1/(r^2) = 92.23
r = 0.104m

However, it's not one of the option...

 

[haruspex]

Homework Statement: At what distance from the 4 x 10-6 C charge is the 2.30 gram mass shown below (file attached) in equilibrium?

The angle of the frictionless incline is 28 degrees.

a) 1.31cm
b) 4.51m
c) 1.60m
d) 0.451m
Relevant Equations: Fg=mgcos(theta)
Fc=kq1q2/r^2
electrostatic constant (k) = 8.99*10^9

View attachment 331275
Fg = Fc
Fg = 2.3g*9.8m/s^2*cos(28)=19.90N
19.90 = (8.99*10^9)*(4*10^-6)*(6*10^-6)/(r^2)
1/(r^2) = 92.23
r = 0.104m

However, it's not one of the option...

A couple of mistakes.
The downslope component of the gravitational force is not mgcos(theta).
Grams are not kilograms.
There may be more.

 

[DanielGuh]

A couple of mistakes.
The downslope component of the gravitational force is not mgcos(theta).
Grams are not kilograms.
There may be more.

Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.

 

[PeroK]

However, I don't know why do we use sin instead of cos.

Why did you use $\cos$ in the first place?

 

[DeBangis21]

Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.

Sin is for vertical axis, cos for horizontal.

 

[PeroK]

Sin is for vertical axis, cos for horizontal.

That doesn't sound right.

 

[Steve4Physics]

Sin is for vertical axis, cos for horizontal.

It doesn't always work like that. In this problem, the direction is the slope's - not vertical or horizontal.

Being able to resolve a force (or any vector) into components is necessary for many sorts of problems. You can click here for a video explaining it from the basics. The last part of the video covers how to deal with an object on a slope.

 

[haruspex]

Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.

A useful check is to think about the $\theta=0$ case. No downslope force, so you want the function that gives 0 at 0.

 

[DanielGuh]

It doesn't always work like that. In this problem, the direction is the slope's - not vertical or horizontal.

Being able to resolve a force (or any vector) into components is necessary for many sorts of problems. You can click here for a video explaining it from the basics. The last part of the video covers how to deal with an object on a slope.

This video helps a lot, thanks!

 

[DeBangis21]

Thank you very much, @Steve4Physics.