Calculate Force Needed to Tip a Rectangular Prism

[flfgw]

Homework Statement

How much force is required to tip an object?
Given- It is a rectangular prism 1m tall, .5 m wide, 2m long.
The center of gravity is .75m from the ground vertically and otherwise centered.
The force can be applied wherever it is easiest to solve the problem.

Homework Equations

The Attempt at a Solution

I don't even know where to start on this one. In order to tip it i need to move the center of gravity over the rotation point. It would not move in a straight line however. It will follow the path of an arc. This is where I'm stumped. I am not to good with torque... does anyone know how to go about solving this?

 

[PhanthomJay]

The amount of force required depends on where the force is applied, and its direction, and it depends also on the weight of the prism. The heavier the prism, the greater will be the force required to tip it over; and it's easier to tip it if the prism is oriented with its width of 0.5m in the direction of the (horizontally applied) load, and when that load is applied horizonatlly at the very top of the prism. Do you have a specific example in mind?

 

[flfgw]

Thankyou for responding. I don't have a specific example. Ill add the following to the given information, sorry it was incomplete at first.
Mass- 500kg
The prism is oriented with its width of 0.5m in the direction of the (horizontally applied) load.
The force is applied in the middle of the 2m side, .9m from the ground at a 45 degree angle above the horizontal and no angle to the left or right.

 

[PhanthomJay]

Thankyou for responding. I don't have a specific example. Ill add the following to the given information, sorry it was incomplete at first.
Mass- 500kg
The prism is oriented with its width of 0.5m in the direction of the (horizontally applied) load.
The force is applied in the middle of the 2m side, .9m from the ground at a 45 degree angle above the horizontal and no angle to the left or right.

The block (let's call it a 2-dimensional rectangle, because the loading is centrally applied in the 3rd dimension) will become unstable and start to tip about the bottom corner when the torque of the applied force, F, about that corner is equal in magnitude to the torque of the block's weight about that corner. You've got to know a little about torques to find the value of F for this condition. Is this what you are looking for? The block won't actually tip over under this condition, because the upward vertical component of this applied force will produce a counteracting torque once the block starts to rotate, and the block will settle into an equilibrium position looking like the Leaning Tower of Pisa.

 

[flfgw]

Thanks for the advice. Unfortunately, I don't know anything about torque but i guess i'll have to learn !

 

[flfgw]

Okay. So i clicked on the torque hyperlink and it was really helpful. I get stuck however when i have to factor in the weight of the block and the center of mass. Do you know how to do that or where i could look to figure it out?

 

[PhanthomJay]

Okay. So i clicked on the torque hyperlink and it was really helpful. I get stuck however when i have to factor in the weight of the block and the center of mass. Do you know how to do that or where i could look to figure it out?

Yeah, torque (or moment) of a force is just the force times the perpendicular distance from the line of action of the force to the point you're taking torques about. Counterclockwise torques must equal clockwise torques for equilibrium; if cw torques are greater or less than the ccw torques, the object will start to to rotate. In calculating torques, it is often useful to break up the force into its x and y components, and calculate the torques of each. In regard to the torque from the weight , its again just the weight of the prism times the perpendicular distance, d, from its center of mass to the pivot. What would this perpendicular distance, from the line of action of the weight force to the pivot, be? Incidentally, I'm trying to visualize your applied 45 degree force loading from your description; because of that upward component of the applied force, it makes a difference whether the applied force is applied to the left side or the right side of the prism.

 

[flfgw]

Okay, another question. Would it require less force to tip the block if the force is only horizontal? If so, then we can make it just a horizontal force with no vertical component.

 

[PhanthomJay]

Okay, another question. Would it require less force to tip the block if the force is only horizontal? If so, then we can make it just a horizontal force with no vertical component.

Not generally speaking. Depending on the geometry, and point of application of the load, the minimum force might be upward, horizontal, or at some angle in between.

 

[flfgw]

How would i find the angle that would require the minimum force?

 

[PhanthomJay]

How would i find the angle that would require the minimum force?

If you looked up some info on torques, you will find 2 ways of calculating it: The first is the one I mentioned, which is Torque = force times perpendicular from the line of action of that force to the pivot point. The second, is that Torque (T) = force (F) times the distance (position vector,r) from the point of application of the force to the pivot, times the sin of the angle in betwen the force and the position vector (T = Fr sin theta). These are identical formulas in different forms. Now since you want to minimize the force that will give you the required torque, you want therefore to maximize the value of 'sin theta'. At what angle, theta, is sin theta a maximum? This will help to tell you in what direction the force should be applied.

 

[flfgw]

I found this online but it was a website i wouldn't trust. Does it make sense to you. The attachment is a diagram which explains the variables. m=mass, g=gravity

F=(m g cos(theta) b) / (a = b)

 

[PhanthomJay]

I found this online but it was a website i wouldn't trust. Does it make sense to you. The attachment is a diagram which explains the variables. m=mass, g=gravity

F=(m g cos(theta) b) / (a = b)

I don't know from your formula or sketch whether 'a' is the length of the diagonal between the lower corner and the cm, or whether it represents an angle 'a' between the vertical and the diagonal, and did you mean (a + b) rather than (a = b) in your denominator? In any case, the figure shows the cm at the centroid of the rectangle. This is different than your case where the cm is not at the centroid. In general, the minimum force required to overturn the block occurs when the force is perpendicular to the position vector. In the diagram shown, that value should be $$F = mgacos\theta/(a+b)$$, where (a + b) is the length of the diagonal.

 

[flfgw]

Thank you so much. Youve been a great help! Yes i did mean that a is the length of the diagonal between the lower corner and the cm. I also did mean (a+b). I am sorry but even with your help I am still not quite certain how this formula was ascertained. Could you please try to explain it.

 

[PhanthomJay]

Thank you so much. Youve been a great help! Yes i did mean that a is the length of the diagonal between the lower corner and the cm. I also did mean (a+b). I am sorry but even with your help I am still not quite certain how this formula was ascertained. Could you please try to explain it.

OK, let's look specifically at this sample problem, not at your problem statement. By looking at the sketch, you should be able to visualize that the block will ultimately tip about the lower left hand corner (the pivot point, where the normal force betwen the ground and the block will be concentrated at the critical 'tipover' force) when you apply a certain force , F, at the top. The force, F, will produce a counterclockwise torque about the pivot. The block's weight, concentrated at the c.m., will produce a clockwise torque about that pivot. When these two torques are equal, the block will just be on the verge of tipping. Thus , set the two torques equal, and solve for F. From the definition of torque, the torque from the weight, mg, is (weight times perpendicular distance from the line of action of the weight force to the pivot). Thus, using a little trig, it's torque is $$mg(a)cos\theta$$, clockwise . (Alternatively, using the second definition for torque I noted earlier, you get the same result, $$mg(a)sin(\theta -90)$$, which is equivalent to $$mg(a)cos\theta$$). Now in a similar fashion, the torque from the force , F, is $$F(a+b)$$, counterclockwise. Set these 2 torques equal, and solve for F, thus, $$mg(a)cos\theta = F(a+b)$$, or $$F = mg(a)cos\theta/(a+b)$$. Now admittedly, this is the case where F is applied at a right angle to the diagonal, which gives the minimum force required. This might be a bit difficult to see. If for example F were applied horizontally at the top, instead of at an angle, it's value would be larger to give the same torque. Now your problem is a little harder, because the c.m. is not at the center of the rectangle. But use the same approach: and note that the line of action of the the force should be perpendicular to the pivot to get the minimum force required.

 

[flfgw]

Thank you sooooooo much. You have beeen an amazing help. I am not going to lie, this was the first time i used this website and i was very doubtful but you have totally changed my opinion. I can't thank you enough!