Basic Ramp Question: Is Force on Ramp & Pole Increased by √2?

[Samky]

Homework Statement

A block of uniform mass is at rest on a horizontal frictionless surface where the net force is given simply by mg (mass x gravity). Now suppose the surface is inclined at 45 degrees to the horizontal to create a ramp. To maintain static equilibrium a pole of negligible mass is placed parallel to the ramp and between the ground and the mass.

Is the following statement correct? The force on the frictionless ramp is given by (mg)cos\theta while the force on the pole is given by (mg)sin\theta and so although mass and acceleration have not changed the net force of the system has increased by a factor of \sqrt{2} given by (mg)(sin45 + cos45). If this is correct then explain where the extra force comes from.

Homework Equations

Strictly speaking this is not a homework problem, I just happened to wonder today and thought this would be the correct forum.

The Attempt at a Solution

I've literally only had one engineering/physics class so go easy on me :)

To me it seems the statement is correct because the component of mg working down the ramp would be sin45 and similarly the component perpendicular to the ramp would be cos45 (or I'm wrong?) I thought about calculating moments, but for example the normal force is (mg)cos\theta so it seems this "extra force" is already built in so to speak. I'm not sure how to approach this.

 

[NuclearNut]

you need to remember that the nett force is zero when the ramp is horizontal, mg down from gravity, mg up from the ramp.

When the ramp is inclined, mg sin theta acts to move the block down the ramp and the pole exerts an equal and opposite force to keep the block in place, similarly mg cos theta acts normal and towards the surface from gravity and the surface exerts an equal and opposite force to stop the block going through the surface. so overall the nett force is equal.

Hope that helps

 

[SammyS]

Force is a vector. Add the two forces vector-wise. They are at right angles to each other.

 

[Samky]

Ack, the net force is zero in both cases, of course you're right. I didn't express my question (or problem) very well at all.

I guess what I'm wondering is why the magnitude of the forces required to maintain equilibrium aren't equal in both cases. Case one requires only mg and case two requires mg (sin theta + cos theta). My reasoning goes that both cases have the same mass and accel values yet require different amounts of force to maintain equilibrium which suggests to me some force was added or subtracted between the two cases(?)

To put it another way, suppose I wanted to accelerate that block and have a resultant velocity vector "P". It seems to require more force for the inclined block to reach v=P than the block on horizontal surface which doesn't make sense to me.

 

[Samky]

Force is a vector. Add the two forces vector-wise. They are at right angles to each other.

Ok, is this right?

case 1:

mg = mg


case 2:

mg sin theta + mg cos theta = mg sin theta + mg cos theta


If so I'm wondering why the force on one side of the equation for case 1 doesn't equal the force for one side of the equation in case 2 considering mass and accel are equal in both cases and f=ma. My reasoning is if ma has not changed from case 1 to case 2 then neither has f although obviously when I plug in numbers this is not the case.

 

[SammyS]

But the force with magnitude mg sin(θ) is perpendicular to the force with magnitude mg cos(θ). You don't add their magnitudes. They're vectors. The square of the magnitude of the resultant is equal to the sum of the squares of the magnitudes of these two vectors.

 

[Samky]

But the force with magnitude mg sin(θ) is perpendicular to the force with magnitude mg cos(θ). You don't add their magnitudes. They're vectors. The square of the magnitude of the resultant is equal to the sum of the squares of the magnitudes of these two vectors.

lol, and sin^2 theta + cos^2 theta is one...

ok ok, I see my error in thinking, thanks for taking the time to help :) This error is kind of embarrassing, oh well glad I see it.

 

[SammyS]

Glad that you came up with it yourself.

You will probably remember it much better than if I had told you the result earlier.

 

[Samky]

Glad that you came up with it yourself.

You will probably remember it much better than if I had told you the result earlier.

This place is awesome

It encourages me to tackle some of the harder problems out of my book this summer if I can come here and have someone point me in the right direction without posting the problem fully worked out (which as you said isn't nearly as helpful).