A couple of mistakes.DanielGuh said:
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.haruspex said:
Why did you use ##\cos## in the first place?DanielGuh said:
Sin is for vertical axis, cos for horizontal.DanielGuh said:
That doesn't sound right.DeBangis21 said:
It doesn't always work like that. In this problem, the direction is the slope's - not vertical or horizontal.DeBangis21 said:
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 said:
This video helps a lot, thanks!Steve4Physics said:
Coulomb's Law describes the electrostatic force between two charged objects. It states that the magnitude of the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. In the context of a charge repelling a mass on a frictionless incline, Coulomb's Law helps determine the electrostatic force acting on the mass due to the charge, which affects the mass's motion up the incline.
The electrostatic force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's Law: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)). On a frictionless incline, this force will act along the line joining the two charges.
The incline angle \( \theta \) affects the components of forces acting on the mass. The electrostatic force has to be resolved into components parallel and perpendicular to the incline. The component of the electrostatic force parallel to the incline will affect the acceleration of the mass up or down the incline, while the perpendicular component is balanced by the normal force from the incline surface.
The mass of the object determines its inertia, which affects how it responds to the electrostatic force. According to Newton's second law, the acceleration \( a \) of the mass \( m \) is given by \( a = F_{\text{parallel}} / m \), where \( F_{\text{parallel}} \) is the component of the electrostatic force parallel to the incline. A larger mass will result in a smaller acceleration for the same force.
The equilibrium position is found when the net force acting on the mass along the incline is zero. This occurs when the component of the gravitational force down the incline \( mg \sin \theta \) is balanced by the component of the electrostatic repulsive force up the incline. Mathematically, \( mg \sin \theta = k \frac{|q_