Bolted Beam Problem: Tension & Force Calculations

[anathema]

These are the kinds of questions that I have the most problems with. Here's the specific question:

"A uniform steel beam of length 5 m has a weight of 4.5x10^3 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25 above the horizontal. A load whose weight is 12x10^3 N is hung from the beam at a point that is 3.5 m from the wall. What is the magnitude of the tension in the supporting cable? What is the magnitude of the force exerted on the ned of the beam by the bolt that attaches the beam to the wall?"

The problem I have is with the bolts, because I don't know what force this is and which direction it has; I initially did the question ignoring the bolts alltogether, until I discovered the second part of the question.

 

[mercury]

what u have to do(i think ) is this..
suppose that the cable did'nt support the beam , then the beam would
feel a torque about the pt. where it is pivoted .ie. the bolt.
so, the toque about that pt. would be
= the moment of inertia of the rod about the pivoted end * angular acceleration

but this torque is balanced by the vertical component of the tension in the cable,ie. T*Sin(25) , equate the downward torque due to gravity and the upward torque due to the tension , take them about the pt. where it is pivoted .(angular acceleration will be g/l ..where l is the length of the rod. )

as to the second part of the question the bolt i think would feel a force of
T*Cos(25) as that is the only other horizontal force..

i think this is right...but I'm only learning this stuff now..so i could very well be wrong...would be glad if anyone could correct it if there's a problem

 

[HallsofIvy]

Well, you can't ignore the bolt or the beam will just dangle from the cable!

The bolt is exerting a force (call it "F") straight up. The cable is exerting a tension force (call it "T") at an angle 25 degrees(?) to the horizontal. The vertical component of the tension force is
T sin(25) so we must have F+ T sin(25)= 16.5x10^3 N, the total weight.

Assuming the weight of the beam is distributed uniformly, we can think of it as concentrated at the center, 2.5 m from the wall. The torque about the wall then is (4.5x10^3)(2.5)+ (12x10^3)(3.5)=
53.25x10^3 N-m. That must be offset by the vertical component of force from the cable: (T sin(25))*5 so
5(sin(25))T= 53.25x10^3. You can find both T and T sin(25) from that. Put T sin(25) back into the first equation to find F.

(There is, by the way, a compression force of T cos(25)N on the beam.)

 

[dg]

Originally posted by anathema
These are the kinds of questions that I have the most problems with. Here's the specific question:

"A uniform steel beam of length 5 m has a weight of 4.5x10^3 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25 above the horizontal. A load whose weight is 12x10^3 N is hung from the beam at a point that is 3.5 m from the wall. What is the magnitude of the tension in the supporting cable? What is the magnitude of the force exerted on the ned of the beam by the bolt that attaches the beam to the wall?"

The problem I have is with the bolts, because I don't know what force this is and which direction it has; I initially did the question ignoring the bolts alltogether, until I discovered the second part of the question.

The answers you have collected so far look pretty euristic (I do not want to go into their correctness) but what I am going to give you is a little more of a universal recipe for this kind of problems.

Your problem is a simple static problem and it can be solved simply by requiring that the (vector) sum of all forces and momenta to be zero (otherwise by Newton's laws there would be linear or angular accelerations and hence motion).

So you start drawing your system and sketching here and there the forces involved in the problem (see word attachment).

Then you have to chose the pole for your momenta calculation. The only smart choice in your case is to pick the point that will eliminate the unknown force at the bolted end (&Omega in the drawing).

At this point you write the equation for zero total force and zero total momentum. Since all your forces and distance vectors lay in a plane you can decompose the forces along two directions (horizontal and vertical) and obtain two (scalar) equations; while all momenta will be orthogonal to this plane and will reduce to one equation.

Momentum equation gives you the tension. Force equations give you each a component of the "bolted end force".

If anything is not clear let me know...