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Forces
Forces  
QUESTION: This is the question I'm trying to answer

A particle of weight W is attached by a light inextensible string of length a to a point A on a vertical wall.  The particle is supported in equilibrium by a light rigid strut of length b attached to a point B on the wall a distance a vertically below A.  Show that the tension in the string is W and find the thrust in the rod.

So, do I have to find the horizontal distance from the wall to the particle and also the vertical distance of the particle to point A in order to solve this problem?  As I'm a beginner in this topic I'm uncertain about thrust.  I am attaching a diagram of what I think I should be working from and would like to know if I'm on the right lines.

ANSWER: Problems in statics like this can be relatively easy to solve by using the key principle of balancing forces being applied to an object. In particular, since the object isn't moving, the component of a force applied in one direction has to have an equal and opposite force acting in the opposite direction. This situation can (and should) be shown in a free-body diagram, pretty much as you have done in your image.

To start, the force affecting everything is gravity, which is pulling down on the weight with a force of W kg. This has to be ballanced by a force pulling up with a force W. This is provided by the upper strut in your diagram. Note, however, that the strut is not vertical so that we need to calculate the vertical component of the tension, T1 (tension is the force along the length of the strut) and make sure it is W. This is where trigonometry comes. The following is the process of writing down various force balances and then solving for the quantities of interest.

Let the angle of the strut from the horizontal be θ. The vertical component of force should then be T1v = T1sinθ = W (we'll figure out angles from the lengths a and b in a second). This also means the horizontal force on the strut pushing it towards the wall, which we can call the horizontal thrust, is T1h = T1cosθ. This horizontal thrust on our object has to be balanced by the lower strut T2. Using φ as the angle of this strut to the horizontal, we have T2h = T2cosφ. This has to be the same as T1cosθ (the horizontal forces are acting in directly opposite directions, of course; I'm using just magnitudes and sort of finessing + and - signs, but you should keep track, at least mentally).

We really want the thrust, T2, along the lower strut. If we know W and the angles θ and φ, we can solve for T2:

T2 = T1・cosθ/cosφ.

Now for the angles: Let the horizontal line from the object to the wall be a length d. This length can be written

d = acosθ = bcosφ

which gives us one eqn for the 2 unknown angles. We can also divide the vertical section of the wall, of length a, between the point of attachment of the 2 struts and the intersection of d with the wall into an upper part of length a1 and a lower part of length a2. Then

a = a1+a2 = asinθ+bsinφ

which is another, indepenedent, eqn relating the 2 angles. We now have the desired situation of having 2 independent eqns in 2 unknowns. I'll leave it 2 you to do the algebra/trig to solve for T2 above.

Let me know if this makes sense.

Randy

---------- FOLLOW-UP ----------

QUESTION: Randy - thanks for the speedy reply.

I must have some misunderstanding of forces,  I'm with idea of balancing forces but I have a problem with the vertical forces.  There is the string "pulling" the weight upwards and it's T1 sin theta but what about the bottom strut's upward thrust?  Why isn't W = T1sin theta + T2 sin fi?

Also in the question it says show that the tension in the string is W.  So if there was no upward thrust from the rod and T1sin theta = W wouldn't theta be 90?

I'm obviously too much of a beginner.  Thanks for taking the time and I hope that you can explain my mis-thinking.

Maggie

Answer
Good follow-up. I did get the vertical forces wrong by forgeting the vertical force along the lower strut. I think the correct force balances should be

vertical:   T1sin(theta) + T2sin(phi) - W = 0

horizontal:  -T1cos(theta) + T2cos(phi) = 0

where the vertical forces are taken as positive if acting in the +y direction and the horizontal forces are positive if in the +x direction.

In particular, W has a minus sign since it is acting in the -y direction. T1sin(theta) has a + sign since it is pulling up, that is, acting in the +y direction. T2sin(phi) must also provide a force in the +y direction, in order to counter the downward force.

For the horizontal balance, there is no horizontal force due to W, so it doesn't appear in the eqn. However, T1 and T2 still need to balance each other. The way I see it, T1 wants to pull the object (point of connection for all 3 components) towards the wall, which would be in the -x direction, hence the minus sign. T2 has to balance this by providing a force in the +x direction.

So, from the 2nd (horizontal) eqn, we have T1 = T2cos(phi)/cos(theta), and substituting into the 1st gives

T2 = W / { [cos(phi)/cos(theta)]sin(theta) + sin(phi) }.

The last bit of sign flipping would be to say that the force T2 in the lower strut is directed out from its end rather than a pulling in, so it can be considered a thrust rather than a tension.

Sorry about the first attempt. My bad. Glad to see you are working through the problem. You're better at it than you think.

Randy

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randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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