a crate on a frictionaless plank inclined at angle theta with respect to the horizontal. Which of the follwing relationships is true? (assume that the x-axis is parallel to the surface of the incline)
a)Fy = Fg
b)Fx=0
c)Fy=Fx
d)none of the above
*my answer would be c but i need to make sure ASAP
thx
force gravity is not in the y direction
Fx is the force down the plank, and without friction, the object is accelerating, and fx=ma, which is not zero
Fy cannot generally be Fx
Your answer doesn't make sense.
The answerwould be C.
The correct relationship in this situation is (b) Fx = 0.
When a crate is on a frictionless inclined plane, the only forces acting on the crate are its weight (Fg) and the normal force (Fn), which acts perpendicular to the surface of the incline. The force of gravity can be resolved into two components, one parallel to the incline (Fg_parallel) and one perpendicular to the incline (Fg_perpendicular).
The normal force (Fn) cancels out the perpendicular component of the weight (Fg_perpendicular), meaning that Fy (the sum of all the forces in the y-direction) is equal to zero. Therefore, Fy = 0.
The only force component remaining is the parallel component of the weight, Fg_parallel, which acts on the crate along the x-axis. As there are no other forces acting in the x-direction, Fx (the sum of all the forces in the x-direction) will also be equal to Fg_parallel. Thus, Fx = Fg_parallel = 0.
Therefore, the correct answer is (b) Fx = 0.
To determine the correct relationship between the forces on the crate on a frictionless inclined plank, we need to consider the forces acting on the crate.
Let's break down the forces involved:
1. The gravitational force (Fg) acts vertically downwards, pulling the crate towards the center of the Earth.
2. The normal force (Fn) acts perpendicular to the plane and prevents the crate from sinking into the inclined plank.
3. The weight component parallel to the incline (Fg_parallel) pulls the crate down the incline.
4. The weight component perpendicular to the incline (Fg_perpendicular) presses the crate into the inclined plane.
Now, let's analyze the forces in the x and y directions:
In the x-direction:
- The weight component parallel to the incline (Fg_parallel) acts downwards, causing acceleration along the x-axis. Therefore, there is a net force acting on the crate in the x-direction.
- The only other force acting in the x-direction is the force of friction (Ff), but since the plank is assumed to be frictionless, Ff is equal to zero.
Hence, Fx = 0 (Option b).
In the y-direction:
- The weight component perpendicular to the incline (Fg_perpendicular) acts downwards, pressing the crate into the inclined plane.
- The normal force (Fn) acts perpendicular to the incline and counterbalances Fg_perpendicular.
- There is no other force acting in the y-direction, so the net force in the y-direction is zero.
Hence, Fy = 0 (Option a).
Therefore, neither option a nor option b is correct.
Since we have established that Fx = 0 and Fy = 0, the correct relationship between the forces is Fy = Fx (Option c).