How to calculate the supporting force of the object on the incli

2. According to the principle that the acting force and the reaction force are equal, the supporting force is equal to the pressure exerted by the object on the inclined plane, that is, f = n = mgcos α

3. Suppose the slope is low on the left and high on the right, and the slope angle is α°， There are three forces acting on the object:

1, gravity, the size of Mg, the direction is downward
2, the supporting force of the inclined plane is perpendicular to the inclined plane (i.e. upward to left) α°
3, the direction of friction is upward along the inclined plane, that is, upward to right (90- α°

since the object is stationary on the inclined plane, the above three forces are balanced

the resultant force of gravity and supporting force is equal to the friction force, and the direction is opposite, that is, down the inclined plane, or down to the left α°， Its size is:
mgsin α°

4. In the specific topic, of course, it can be found out. It should be noted that whether the inclined plane is fixed or not, the supporting force on the inclined plane is always perpendicular to the inclined plane.

5. The supporting force on the object is equal to the pressure on the inclined plane. Because the gravity of the object is divided into the pressure on the inclined plane and the downward friction! Center of gravity ~!

6. 1. No work (no friction)
2. W = FS = 20n × 10M=200J
3.W=Fs=100N × 1m = 100J
4. No work done (the moving distance of the working object in the direction of the force)

almost. Haha

thank you for your adoption

7. Orthogonal decomposition. The x-axis is built along the inclined plane and the y-axis is built perpendicular to the inclined plane. The resultant force is zero in the y-axis direction. Thus, FN = mgcos 30 degrees. F = mgcos 30 degree
all questions can be solved according to this idea
in addition, according to your intention, if an object is stationary or moving in a straight line at a constant speed, you can directly use the equal quadrilateral rule to make a parallelogram with any two forces as the adjacent sides. Its diagonal is equal to three equal forces. It can also be solved. Take F and G as adjacent sides to make a parallelogram, and the size of the diagonal is equal to FN. FN = = mgcos 30 degrees can also be obtained. F = mgcos 30 degrees


G

8. G = Mg (the gravity on the slope is the same as that on the plane)

9. Sliding force? It's the first time I've heard that the support force is equal to mgcosx, which is the component force of gravity perpendicular to the inclined plane.
centrifugal force has nothing to do with gravity. Centrifugal force is a virtual force, not a real one. For an object moving in a uniform circular motion, the centrifugal force and centripetal force are equal in size and opposite in direction. Centripetal force is the force that makes the object move in a circular motion, The centripetal force on an object moving in a uniform circular motion = MV ^ 2 / R
again, centrifugal force is not a real force, but a virtual thing

10. There is no supporting force at all, because support belongs to the category of elastic force, and one of the conditions for elastic force is that it must be contacted,
there is no contact between the ground and the object, so the ground does not give the object supporting force