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Forces Concerning a Rectangular Prism 
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#1
Jan313, 08:30 PM

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1. The problem statement, all variables and given/known data
I'm wondering this for any object with moment of inertia I, but I'll ask this question for a rectangle for simplicity and I'm sure I can extend it to general objects. Say we have a rectangular object (with mass m, height h, and width L) that is attached to a wall by some sort of point support exactly at its corner. If we hold it horizontally then let it go and swing, what is the force that the support exerts on the ruler from right after we let go as it swings? Here's a picture: 2. Relevant equations Moment of inertia of this rectangle: [itex]\frac{m(h^2 + w^2)}{12} + m((h/2)^2+(w/2)^2) = \frac{m(h^2+w^2)}{3}[/itex] 3. The attempt at a solution I was thinking about concentrating the mass into a point mass at the location of the rectangle's center of mass. This would be like a simple pendulum with starting position at angle [itex]θ[/itex] below the horizontal, where [itex]θ=\tan^{1}(h/w)[/itex]. There would be a force of mg downward, which has a component of mg*sin(θ) in the direction of swinging and a component of mg*cos(θ) pulling on the "string" of the pendulum. The force that the support supplies, in return, would be opposite to and equal in magnitude to this pull. I'm not so sure this is correct though... 


#2
Jan313, 11:47 PM

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Try calculating the torque around the pivot point.



#3
Jan313, 11:53 PM

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How does this help me find the force exerted by the support at the pivot point though? 


#4
Jan413, 06:59 AM

P: 5,462

Forces Concerning a Rectangular Prism
You say it has moment of inertia I so do you understand the calculation of moments of inertia? Nugatory has offered a method that uses the same approach. 


#5
Jan413, 12:11 PM

P: 15

Woops, [itex]θ=\tan^{1}(w/h)[/itex], not [itex]h/w[/itex].
With the moment of inertia, I can figure out the motion of this rectangle. How does that relate to the force that the pivot supplies? Does it supply a force equal to the component of the force of gravity on the rectangle, that is perpendicular to the component in the direction of the swinging motion? 


#6
Jan413, 03:56 PM

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Calculating the torque around the pivot point also works. That gives you the angular acceleration, and the linear acceleration to add to that must be that which holds the pivot point still. Then you can figure out what force at the pivot point when combined with gravity produces that linear acceleration. The two methods are equivalent. 


#7
Jan413, 03:58 PM

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#8
Jan413, 11:59 PM

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#9
Jan513, 12:58 AM

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The two aspects of movement result from two forces, gravity and the force at the pivot, acting together. I didn't really understand your question, so it's hard to guess whether I've answered it. 


#10
Jan513, 01:17 AM

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#11
Jan513, 09:27 AM

P: 358

You are basically considering a physical pendulum here. You do have circular motion of the cm about the pivot point. It seems you are particularly interested in the radial acceleration of the pendulum which will then enable you to evaluate the force.



#12
Jan513, 12:44 PM

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#13
Jan513, 02:46 PM

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the force from the pivot will also have a component perpendicular to that radial line. Together with the corresponding component from gravity this will provide the torque necessary for the angular acceleration.I feel it will become clearer if you just go ahead and develop the usual equations. 


#14
Jan513, 04:02 PM

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The torque due to the component of gravity perpendicular to the radius is [itex]mgL\sin{θ}[/itex]. Since [itex]\sin{θ}=\frac{w}{\sqrt{w^2+h^2}}[/itex] and [itex]L=\frac{\sqrt{w^2+h^2}}{2}[/itex], the torque is therefore [itex]\frac{mgw}{2}[/itex]. This torque is equal to [itex]Iα[/itex] and [itex]I=\frac{m(w^2+h^2)}{3}[/itex]. So, [itex]α=\frac{3gw}{2(w^2+h^2)}[/itex]. [itex]a = αL = \frac{3gw}{2(w^2+h^2)}*\frac{\sqrt{w^2+h^2}}{2}[/itex] [itex]=\frac{3gw}{4\sqrt{w^2+h^2}}[/itex] The acceleration from gravity, [itex]g\sin{θ}=\frac{gw}{\sqrt{w^2+h^2}}[/itex] is larger than [itex]a[/itex] found just above! So, I'm thinking that the pivot supplies a force in the opposite direction to make the net acceleration [itex]α=\frac{3gw}{4(w^2+h^2)}[/itex] This force is [itex]m(\frac{gw}{\sqrt{w^2+h^2}}\frac{3gw}{4(w^2+h^2)})=\frac{mgw}{4\sqrt{w^2+h^2}}[/itex] So, my conclusion is that the force from the pivot is in two components: the first is in the radial direction (which makes the radial net force 0). This force has a magnitude of [itex]\frac{mgh}{\sqrt{h^2+w^2}}[/itex], which is the radial component of force due to gravity. The other component is perpendicular to the previous component. This was found to be [itex]\frac{mgw}{4\sqrt{w^2+h^2}}[/itex]. Is my thinking correct? 


#15
Jan513, 04:27 PM

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It's easier just to say the line of action of mg is distance w/2 from the pivot, so the moment is mgw/2. 


#16
Jan513, 05:10 PM

P: 15

Oh, good catch. No wonder the rest of my problem wasn't working out. Thanks!



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