Physics Help Forum Inclined plane

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 Aug 14th 2008, 09:43 PM #1 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 Inclined plane Why the answer is not zero but 24N? For force applied F, where comes a force making the wedge stationary? What is the effect of F on block? (Is the force parallel to ground or perpendicular to the inclined pland? What is the magnitude of F on block?) Attached Thumbnails
Aug 15th 2008, 07:44 AM   #2
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Hello,
 Originally Posted by werehk Why the answer is not zero but 24N?

There are four forces acting on the wedge : the force exerted by the ground $\displaystyle \vec{N}_{\text{ground}}$, its weight $\displaystyle \vec{W}_{\text{wedge}}$, the force $\displaystyle \vec{F}$ and the force $\displaystyle \vec{N}$ exerted by the block. We are told that the wedge is at rest so, using Newton's second law :

$\displaystyle \vec{N} + \vec{W}_{wedge} + \vec{N}_{\text{ground}} +\vec{F}=\vec{0}$

As "all contact surfaces are assumed to be smooth", the force exerted on the wedge by the ground is vertical so the projection of the previous relation on the x-axis (horizontal) gives us

$\displaystyle \vec{N}\cdot \vec{x} + \underbrace{\vec{W}_{wedge}\cdot \vec{x}}_0 + \underbrace{\vec{N}_{\text{ground}}\cdot \vec{x}}_0 +\underbrace{\vec{F}\cdot \vec{x}}_{\pm\|\vec{F}\|}=\vec{0}$

hence

$\displaystyle \|\vec{F}\|=|\vec{N}\cdot \vec{x}|=\| \vec{N}\|\times \underbrace{\|\vec{x}\|}_1 \times |\cos \theta_1| \implies \boxed{\|\vec{F}\|=\| \vec{N}\| \times |\cos \theta_1|}\,\,\,\,\,\,\,(1)$

Now, what is $\displaystyle \vec{N}$ ? Let's take a look at the attached picture. The two forces which are acting on the block are its weight $\displaystyle \vec{W}=m\vec{g}$ and the force exerted by the wedge which is $\displaystyle -\vec{N}$. Again, as "all the surfaces are assumed to be smooth", we get $\displaystyle \vec{N}\perp (AC)$. When the block slides, we know it moves along a straight line parallel to $\displaystyle (AC)$ so the direction of its acceleration is parallel to $\displaystyle (AC)$ too. Using Newton's second law we get

$\displaystyle m\vec{a} =\vec{W} -\vec{N}=m \vec{g}-\vec{N}$

The projection of this relation on an axis $\displaystyle z$ perpendicular to $\displaystyle (AC)$ gives us

$\displaystyle \underbrace{m\vec{a}\cdot \vec{z}}_0=\vec{W}\cdot \vec{z}-\underbrace{\vec{N}\cdot \vec{z}}_{\pm\|\vec{N}\|} \implies \boxed{\|\vec{N}\|=|mg\cos \theta_0|}\,\,\,\,\,\,\,(2)$

Putting this together with (1) :

$\displaystyle \|\vec{F}\| =| mg\cos \theta_0\cos \theta_1|=5 \times 9.81\times \frac{40}{\sqrt{40^2+30^2}}\times \frac{30}{\sqrt{40^2+30^2}}=23.5\,\text{N}$

 For force applied F, where comes a force making the wedge stationary?
I'm not sure to understand what you're asking. If you want to know how can one exert the force $\displaystyle \vec{F}$ on the wedge then one solution is to put the wedge against a wall, the wall being on the right side of the wedge (any other object fixed to the ground will do).
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Last edited by flyingsquirrel; Aug 15th 2008 at 08:00 AM.

Aug 17th 2008, 05:24 PM   #3

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 Originally Posted by werehk Why the answer is not zero but 24N? For force applied F, where comes a force making the wedge stationary? What is the effect of F on block? (Is the force parallel to ground or perpendicular to the inclined pland? What is the magnitude of F on block?)
Hint: F will be equal to the horizontal force the block exerts on the wedge. So resolve the normal force into components into the horizontal and vertical directions.

-Dan
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