Finding coefficient of friction and tension in a steel beam?

Question

Steel construction beams, with an industry designation of "W12 multiplied by 22," have a weight of 20 pounds per foot. A new business in town has hired you to place its sign on a 4.0 m long steel beam of this type. The design calls for the beam to extend outward horizontally from the front brick wall (see figure). It is to be held in place by a 5.0 m-long steel cable. The cable is attached to one end of the beam and to the wall above the point at which the beam is in contact with the wall. During the initial stage of construction, the beam is not to be bolted to the wall, but to be held in place solely by friction.

Picture to go with problem: http://i284.photobucket.com/albums/ll28/bathtub2008/12-50.gif

(a) What is the minimum coefficient of friction between the beam and the wall for the beam to remain in static equilibrium?

_____

(b) What is the tension in the cable in this case?

_____N

Answer 1

To answer this question properly, you need to develop a free body diagram and find your resultant forces.

First off, I would like to note that a W12x22 should weigh 22 lbs/ft, not 20.  The second number in the wide flange designation corresponds to lbs/ft weight, where as the first corresponds to the depth of the beam.  Refer to AISC manual.  However, I will use 20 lbs/ft in this calculation.

First, converting the weight of the beam to metric units, we have that it will weigh:

20lbs/ft(1.48816394kg/m/lb/ft)(9.81m/s... = 1167.911 N

This is the total weight of the beam.  Since it is a uniform beam, we know that this total weight can be shown as just a central point load for our free body diagram.

For the FBD, there will be the weight at the CG, a load acting in the direction of the cable, as well as a horizontal and vertical force at the wall.  The vertical force is the friction force.

Summing moments around the point where the cable is attached to the beam, we get:

Ff(4)=W(2)

Ff=0.5(1167.911) = 583.9556 N upwards

Note that Ff is the frictional force and W is the weight of the beam.

Now knowing that the beam is in equalibrium, all the vertical forces must sum to zero.

Ff + W + Ty = 0

Ty = 1167.911 - 583.9556 = 583.9556 N upwards.

Note Ty is the vertical force component of the cable tension.

Now we need to determine the angle of the cable, which will be:

theta = arcos(4/5) = 36.8699 degrees

Knowing that the vertical component of tension in the cable is 583.9556 N, we know that the total tension must be equal to:

Tsin(theta) = Ty

T = Ty/sin(36.8699) = 583.9556/0.6 = 973.2592 N

This is the total tension in the cable.

Now we need to solve for the normal force at the wall, which will be found by knowing the total horizontal forces in the beam will sum to 0.

Tx = Fn

Fn = Tcos(36.8699)

Fn = 973.2592(0.8) = 778.6074 N

The static coefficient of friction is equal to:

u = Ff/Fn

u = 583.9556/778.6074 = 0.75

Note that static coefficient of frictions usually are in the 0.2 - 0.4 range for steel, so chances are that the beam will fall.