On an inclined plane is Gravitation Force of x component (Fgx) always equal to static friction (fs)?
not if the thing moves !!!!
Angle to horizontal = A
weight = m g
component of weight down slope m g sin A
normal force = m g cos A
friction force up slope = mu m g cos A = static friction force up slope (fs)
your Fgx = m g sin A
Fgx - fs = m a where a is acceleration down the slope. (use moving, not static friction once it starts)
Well, wouldn't that be a "slippery" situation? The gravitational force of the x component (Fgx) and static friction (fs) are actually two different forces, so they don't have to be equal. The static friction opposes the motion or tendency of motion on an inclined plane, while the gravitational force contributes to the object's weight. So, in general, Fgx and fs will not be equal. However, it's worth mentioning that there can be a case when Fgx and fs are equal, but it's more of an exceptional circumstance rather than the norm.
No, the gravitational force along the x-axis (Fgx) is not always equal to the static friction force (fs) on an inclined plane. The forces involved are dependent on several factors, including the angle of the inclined plane, the mass of the object, and the coefficient of friction.
The force of gravity acting on an object can be resolved into two components when the plane is inclined: the force of gravity along the direction of the plane (Fgx) and the force of gravity perpendicular to the plane (Fgy).
The static friction force (fs) opposes the tendency of the object to slide down the inclined plane. It is determined by the coefficient of static friction (μs) and the perpendicular force (Fgy) acting on the object. The formula for static friction is fs = μs * Fgy.
Therefore, the gravitational force along the x-axis (Fgx) is equal to fs only when there is no relative motion between the object and the inclined plane. If the object is stationary or moving with a constant velocity, Fgx will be equal to fs. However, if the object is accelerating or sliding down the plane, Fgx will be greater than fs.
To determine whether the Gravitational Force x-component (Fgx) is always equal to static friction (fs) on an inclined plane, we need to consider the forces acting on the object.
On an inclined plane, the two main forces at play are the gravitational force and the normal force. The gravitational force, Fg, acts vertically downward and can be broken down into two components: Fgx (the force pulling the object parallel to the plane) and Fgy (the force pulling the object perpendicular to the plane).
The normal force, Fn, acts perpendicular to the plane and counteracts the gravitational force in the y-direction.
When an object is at rest on an inclined plane, the static friction force, fs, comes into play. It acts parallel to the surface and opposes the tendency of the object to slide down.
Now, comparing Fgx with fs:
If the object is not moving, meaning it is in a state of static equilibrium, then fs will be equal in magnitude and opposite in direction to Fgx. This ensures that the object doesn't slide down the plane.
However, it is important to note that fs can vary depending on the angle of the incline, the coefficient of friction between the surfaces, and other factors. It is not always the same as Fgx, but it needs to be equal in magnitude and opposite in direction to maintain static equilibrium.
To calculate the magnitude of static friction (fs), you generally need information about the coefficient of static friction (μs) between the surfaces in contact. The formula for calculating fs is: fs = μs * Fn, where Fn is the magnitude of the normal force.
In conclusion, the Gravitational Force x-component (Fgx) is not always equal to static friction (fs) on an inclined plane, but fs needs to be equal in magnitude and opposite in direction to Fgx to prevent the object from sliding down the plane.