User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 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?)   Aug 15th 2008, 07:44 AM   #2
Junior Member

Join Date: May 2008
Posts: 17
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).

Last edited by flyingsquirrel; Aug 15th 2008 at 08:00 AM.   Aug 17th 2008, 05:24 PM   #3

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,814
 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.  Tags inclined, plane Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post alagarn Energy and Work 1 Mar 26th 2012 11:07 AM mess@physics Thermodynamics and Fluid Mechanics 2 Dec 1st 2010 01:17 PM nmyers77 Kinematics and Dynamics 0 Nov 21st 2010 08:46 PM subho Kinematics and Dynamics 3 Sep 2nd 2010 12:17 AM ignus275 Kinematics and Dynamics 7 Mar 22nd 2009 08:01 AM