What is the force exerted by a ramp on a mass?

[Alex_Neof]

Homework Statement

Consider a ramp at an angle of 30° to the horizontal with a mass m sitting on it without sliding:

What is the force exerted by the ramp on the mass? Then resolve this force into vertical and horizontal components.

Homework Equations

Net forces equating to zero.

The Attempt at a Solution

The normal force is the force exerted by the ramp on the mass sitting on the ramp.
i.e. F = mg*cos(30) = mg*sqrt(3)/2

Then resolving this force into horizontal and vertical components:

F(vert) = (mg*sqrt(3)/2)*sqrt(3)/2 = (3mg)/4
F(hor) = (mg*sqrt(3)/2)*sin(30) = mg*sqrt(3)/4

I believe this to be correct, however the answer given, states that the force exerted by the ramp in mg, and thus the vertical component of this force is mg and horizontal component of this force is 0.

But which is correct?

 

[PhanthomJay]

Homework Statement

Consider a ramp at an angle of 30° to the horizontal with a mass m sitting on it without sliding:

What is the force exerted by the ramp on the mass? Then resolve this force into vertical and horizontal components.

Homework Equations

Net forces equating to zero.

The Attempt at a Solution

The normal force is the force exerted by the ramp on the mass sitting on the ramp.
i.e. F = mg*cos(30) = mg*sqrt(3)/2

Then resolving this force into horizontal and vertical components:

F(vert) = (mg*sqrt(3)/2)*sqrt(3)/2 = (3mg)/4
F(hor) = (mg*sqrt(3)/2)*sin(30) = mg*sqrt(3)/4

I believe this to be correct, however the answer given, states that the force exerted by the ramp in mg, and thus the vertical component of this force is mg and horizontal component of this force is 0.

But which is correct?

You did not consider all forces exerted by the ramp on the mass. The normal force is one of these forces, are there others?

 

[Alex_Neof]

oh right, so is the force exerted by the ramp on the mass the magnitude of the result force of both the frictional force = mg*sin(30) and the normal force = mg*cos(30)?

So resultant Force = Sqrt[(mg*sin(30))^2 + (mg*cos(30))^2] = Sqrt[(mg)^2 (sin^2(30) + cos^2(30))] = Sqrt[(mg)^2] = mg.

 

[PhanthomJay]

oh right, so is the force exerted by the ramp on the mass the magnitude of the result force of both the frictional force = mg*sin(30) and the normal force = mg*cos(30)?

So resultant Force = Sqrt[(mg*sin(30))^2 + (mg*cos(30))^2] = Sqrt[(mg)^2 (sin^2(30) + cos^2(30))] = Sqrt[(mg)^2] = mg.

Yes, and the angle of that resultant force of the ramp on the mass with respect to the vertical is__??

 

[Alex_Neof]

0 degrees with respect to the vertical, since the resultant force is mg, which is exactly opposite the object's weight = mg. ∑F(vert) = 0.

 

[Alex_Neof]

Thank you PhanthomJay. Kind Regards!

 

[PhanthomJay]

Thank you PhanthomJay. Kind Regards!

You are welcome. This is good work and sound reasoning on your part.