What is the distance HP?

Question

A uniform beam HK of length 10k and weighing 200N is supported at both ends. A man weighing 1000N stands at a point P on the beam. If the reactions at Holborn and K are respectively 800N and 400N. What is the distance HP.

Answer 1

Based on my interpretation of the information, I would set the problem up like this:

 

x = distance HP from H to the 1000 N man

 

Put the 200 N weight of the beam through the center of mass at 5m (length = 10 m)

 

In clockwise direction, the net torque about point H must be zero to be in equilibrium:

 

τ_H = x(1000 N) + (5m)(200 N) - (10 m)(400 N) = 0

 

(Note the force applied at point H does not contribute to the torque about H.)

 

x = (10*400 - 5*200) / 1000 m = 3 m

 

We can check this by calculating torque about K in the same way:

 

(800 N)(10 m) - (200 N)(5 m) - (1000 N)(10m - x) = 0

 

10 - x = (800*10 -200*5) / 1000 = 7m

 

Then x = 10-7 m = 3m, so it checks out.

 

 

 

Answer 2

Hi JULCIT!

 

The solution to this lies in setting up one of the conditions for static equilibrium:  the fact that the sum of torques must be zero.  In the above statement of the problem, I am assuming the length of the is 10 m (if it is 10 km, the same principles would apply... and that would be one long beam!).  I am also assuming the beam is horizontal, which is usually the case for this kind of problem, and that it is being supported at both ends by vertical normal forces exerted by some supports underneath the beam (this could be changed to vertical ropes holding up the beam with no effective change to the solution).

 

Torque can be thought of as the effectiveness of a force in causing rotation.  For a horizontal beam on which vertical forces are acting, the expression for the magnitude torque produced by each force F is:

 

|τ| = Fr

 

where r is the offset from some pre-defined axis of rotation (the distance from the axis to where the force is being applied).  [NOTE:  this expression would have to be modified if the beam were not oriented perpendicular to all the forces applied to it]  The nice thing about this situation is that we can choose the axis of rotation to be wherever we want.  So choosing a most convenient position for the axis would be helpful.

 

In this case, we want the answer as a distance measured from Point H.  So let's choose Point H as the axis of rotation.  We can then determine the magnitude of each torque based on measuring its offset r value from Point H.

 

In cases like this, we typically talk about torques that can make an object rotate either clockwise or counterclockwise.  This is just like one-dimensional motion along, for example, the x-axis or the y-axis, where you can either go one way or the other.  So we can signify these tow directions by + and -.  Conventionally, torques that would, by themselves, cause counterclockwise rotation around the chosen axis point are "plus," and torques that would, by themselves, cause clockwise rotations around the chosen axis point are "minus."

 

With this in mind, let us calculate the magnitude and direction of all the torques acting around Point H.  If the object is in equilibrium, these torques should add up to zero.  We will look at each force in turn (this would be perhaps easier to follow if we could have a diagram; I apologize):

 

1.  The normal force of support at Point H

This force pushes (I assume) vertically up on the board, with a magnitude of 800 N (I assume the "reaction at Holborn" refers to the normal force at Point H).  Since this force is applied at Point H, the distance from Point H to where the force is being applied -- a distance I will denote as r_h -- is zero.  Hence, the magnitude of the torque it produces is:

 

|τ_h| = (800 N)(0 m) = 0

 

This force produces no torque around this chose axis of rotation, because it has no offset.  It is applied directly at the axis.  This is like trying to open a door by pulling on the hinges.  You can apply a big force; but, with no offset from the axis of rotation, it will not make the door rotate.  In other words, it produces no torque.  This is why torque can be thought of as the effectiveness of a force is causing rotation.  You can have as large a force as you want, but -- if it has no offset -- it will produce no rotation, and will thus exert no torque.

 

 

2.  The weight of the beam

The magnitude of this force is given as 200 N.  But where is it applied to the beam?  Torque is often the first physical quantity encountered by students where the position at which the force is applied matters as much as (or more than) the magnitude of the force.  When dealing with Newton's laws, and work and energy, all that usually mattered was the magnitude of the force.  If you drew a box, and wanted to represent its weight, exactly where you drew the weight force arrow from was not important.  But, for torque, position matters, because torque depends on force magnitude AND offset.

 

For this purpose, we typically define a location called the center of mass (or, for the purpose of locating where weight is applied, the center of gravity), which is defined as the point where we can consider any external force as being applied to the object for the purpose of calculating its net acceleration.  There is a standard formula for calculating that; but, in this case, we are given a shortcut for that, because we are told the beam is uniform.  For a uniform object, the center of mass is located at its geometric centroid.  For an object with symmetry, like a beam (which is basically one-dimensional, for our purposes), the centroid is at the center of symmetry.  In this case, that is the midpoint of the beam.

 

Hence, the weight is applied 5 m away from Point H, at the center of the beam.  The magnitude of the torque it produces is thus:

 

|τ_w |= (200 N)(5 m) = 1000 N·m

 

Now, to assign the proper direction sign.  I drew a diagram of the situation, with Point H on the left of the beam.  In this case, you might see that, acting all by itself (as if none of the other torques were present), the weight would cause the beam to rotate clockwise around Point H (as the beam fell).  So I will put a - sign in front of this torque, meaning:

 

τw = -1000 N·m

 

3.  The man's weight

The magnitude of the man's weight is 1000 N, and it is applied at Point P, a distance HP (which we are trying to solve for) from Point H.  In the way I described drawing it above, the man's weight would also -- acting by itself -- cause the beam to rotate clockwise around Point H.  Hence, the torque of the man's weight (so far as we can write it now) is:

 

τ_m = -(1000 N)(HP) = -1000(HP) N·m

 

4.  The support at Point K

The final force that could exert a torque on the beam is the support at Point K.  We assume again this is a vertical normal force, and we are told its magnitude is 400 N.  Its offset is just the distance from Point H to Point K; in other words, the length of the beam, which I assume (as I mentioned) to be 10 m.  The way I have drawn, this force, acting alone, would cause the beam to rotate counterclockwise around Point H, so we can write its torque as:

 

τ_P = (+)(400 N)(10 m) = 4000 N·m

 

Now, we have all the torques on the board, and the equilibrium condition is that they must add up to zero.  This means:

 

τ_net = τ_H + τ_w + τ_m + τ_P = 0

 

Putting in our torque values (including sign) from above, this becomes:

 

0 - 1000 N·m - 1000(HP) N·m + 4000 N·m = 0

 

This is one equation with one unknown:  HP, the distance we are trying to solve for.

 

[NOTE: if I had drawn a diagram with Point H on the right, and the beam extending to the left, I would just change all my + signs to -, and vice versa, which would not change the equation; I would get the same result]

 

Now I will drop the units, for a bit, to make this look a bit more like the algebra problem it is:

 

-1000 - 1000(HP) + 4000 = 0

 

Just solve this expression for HP.  The minimum success criterion, as we like to say, is that HP has to be less than 10 m.  Otherwise the man would be off the board.  

 

 I did get a result for HP that made sense.  If you would like to check an answer, just let me know.

 

[NOTE:  because the support at Point H is providing more upward force than the support at Point K, we might expect that the man must be closer to Point H than Point K, so that the support at Point H would have to take more of his weight.  This would mean HP is less than 5.  My answer met this criterion, too]

 

Hope this helps!