Please clear my doubt about acceleration.

[Shahroze]

In real life if a block slips (in contact always) on a wedge then which force is responsible to maintain the ratio i.e ay/ax = tan (theta). I mean to say that y direction has 'g' but what is in x direction? Theta= angle of inclination! I know if all the surfaces are frictionless and wedge is not fixed then the wedge and block both will slide so ax would be caused by a pseudo force but what about in real life where wedges don't slide?

[Country Boy]

When an object is sitting on the ground there are two forces- the force of gravity pointing downward and the force due to the pressure the ground is exerting on the object.  Those are "equal and opposite", they cancel and one could, as well, say there is no (net) force.  If an object is sitting on a frictionless tilted surface such as a wedge  there is the force of gravity but now the force due to the surface is not directly opposite to gravitational force and we may have motion along the surface.  In "real life", with friction, the friction force is parallel to the surface so we can think of this as three forces- the vertical force of gravity, the pressure force due to the surface which is perpendicular to the surface, and friction which is parallel to the surface.

[swansont]

1 hour ago, Country Boy said:

When an object is sitting on the ground there are two forces- the force of gravity pointing downward and the force due to the pressure the ground is exerting on the object.  Those are "equal and opposite", they cancel and one could, as well, say there is no (net) force. 

A word of caution: The phrase "equal and opposite" suggests Newton's third law, and these are not an example of action/reaction force pairs (which act on different objects, and never show up in an application of the second law)

The force the surface exerts on the body is commonly called the normal force. It is normal (perpendicular) top the surface. When the surface is at an angle, it will have a component in both the x and y directions.

Alternately, you can choose a coordinate system such that the normal force is in the y direction, and gravity would have components in both the x and y directions. Choice of coordinate system will not change the answer (although it may make the solution easier or harder to work out)

 

[Shahroze]

5 hours ago, swansont said:

A word of caution: The phrase "equal and opposite" suggests Newton's third law, and these are not an example of action/reaction force pairs (which act on different objects, and never show up in an application of the second law)

The force the surface exerts on the body is commonly called the normal force. It is normal (perpendicular) top the surface. When the surface is at an angle, it will have a component in both the x and y directions.

Alternately, you can choose a coordinate system such that the normal force is in the y direction, and gravity would have components in both the x and y directions. Choice of coordinate system will not change the answer (although it may make the solution easier or harder to work out)

 

I want to know what will be the acceleration of block in vertical and horizontal direction. Is it gsin^2(theta) and gsin(theta).cos(theta). If this is so then should the force exerted by the block on the wedge be mgsin^2(theta)? Please help i am very confused about the forces as well and why can't it be mgsin^2(theta)

[swansont]

There should be no squared trig functions in the solution. How are you getting them?

The acceleration components will be g sin(theta) horizontally and and g cos(theta) vertically, where theta is the angle of inclination of the block.

Here is the free body diagram of the problem. (In this case we would ignore any applied force, which would include friction, as they are not mentioned in your setup)

http://thecraftycanvas.com/library/online-learning-tools/physics-homework-helpers/incline-force-calculator-problem-solver/

Because we know there is no acceleration perpendicular to the surface of the incline, we can say that N = mg cos(theta)

 

[Shahroze]

5 minutes ago, swansont said:

There should be no squared trig functions in the solution. How are you getting them?

The acceleration components will be g sin(theta) horizontally and and g cos(theta) vertically, where theta is the angle of inclination of the block.

Here is the free body diagram of the problem. (In this case we would ignore any applied force, which would include friction, as they are not mentioned in your setup)

http://thecraftycanvas.com/library/online-learning-tools/physics-homework-helpers/incline-force-calculator-problem-solver/

Because we know there is no acceleration perpendicular to the surface of the incline, we can say that N = mg cos(theta)

 

gsin(theta) horizontally or along the plane? I had read somewhere that acceleration of the block in vertical direction(not along the plane) should be sin^2(theta) if thats true then the force by the block in the vertical direction should be 

7 hours ago, swansont said:

A word of caution: The phrase "equal and opposite" suggests Newton's third law, and these are not an example of action/reaction force pairs (which act on different objects, and never show up in an application of the second law)

The force the surface exerts on the body is commonly called the normal force. It is normal (perpendicular) top the surface. When the surface is at an angle, it will have a component in both the x and y directions.

Alternately, you can choose a coordinate system such that the normal force is in the y direction, and gravity would have components in both the x and y directions. Choice of coordinate system will not change the answer (although it may make the solution easier or harder to work out)

 

I am asking about the vertical acceleration not the Acceleration along the plane! I know acceleration along the plane should be gsin(theta)

[swansont]

 

5 hours ago, Shahroze said:

I am asking about the vertical acceleration not the Acceleration along the plane! I know acceleration along the plane should be gsin(theta)

So take the setup and solve for the vertical. You have an equation that relates N and g, so you can solve and get an answer that depends on g and the angle