Here are some problems for this section that are already worked out for you to examine to help further your understanding before attemptin some on your own. This will also give you an idea of how to apply the concepts that we learned to actual problems.
A bridge that has a mass of 1500 kg and is 6.2 m long is supported by two columns, A and B. If a 3000 kg car as shown in above picture, drives onto the bridge, what force must exerted by columns A and B to support the load if the beam's weight is exerted at its middle?
Since the bridge is not accelerating upwards or downwards, we can conclude that the net force must equal zero. This means we can write the following equation:
Fa + Fb + mg + Fcar = 0
Fa + Fb + -14700 N + -29400 N = 0
Fa = 44100 N - Fb
Mg and Fcar are both negative since they are acting in a downward direction. We also know that since the object is in equilibrium, the net torque also equals zero. Therefore picking the point where force A acts as a pivot point, we can write the following equation:
Torque CW = Torque CCW
Fg * l1 + Fcar * l2 = Fb * l3
14700 N * 3.1 m + 29400 N * 4.8 m = Fb * 6.2 m
186690 N*m = (6.2 m)Fb
Fb = 30111 N
Now that we know the value of Fb, we can sub it back into the orginal equation and solve for Fa.
Fa = 44100 N - 30111 N
Fa = 13989 N
Of course if you were looking a significant figures, you would need to round these answers.
A sign with a mass of 500 g is suspended at the end of a 50 g beam by a rope at the angle shown in the above diagram. If the rope is massless, find the tension in the rope if the beam is 1.0 m long with its weight acting at its center.
First of all, we need to understand all of the forces acting in the example. We can see that both the weight of the beam and the sign are forces acting downwards, and that T is a force acting on an angle with both horizontal and vertical components. Although it is not as obvious, the wall also exerts a force in an angular direction as well. This must be so as the object is in equilibrium and there must be a component of the horizontal net force exerted by the wall in order to balance the system as it is in equilibrium. Since we don't know the value of the force exerted by the wall, we cannot equate all the forces to find the missing values. This means we will have to deal with torques.
Because we have no info concerning the wall's forces in this problem, let's forget about it when we start dealing with torques by placing the pivot point where the wall is and using that spot when calculating the torques. Now to deal with the torques, we need a value for the vertical component of T which we can easily make using trig. The horizontal component of T will not come into play as it has no perpendicular distance from the pivot point and thus does not cause a torque on the system.
Cos30 = Fv/T
Cos30 (T) = Fv
Now that we have a value for each force in the equation, we can calculate the value of T using the laws concerning equilibrium and torques to create an equation. We also have to make sure to put our mass values in Newtons before placing them into the equation.
Torque CW = Torque CCW
4.9 N * 1.0 m + 0.49 N * 0.5 m = T(Cos30) * 1.0 m
5.149 = T(Cos30)
5.9 N = T
Now that you have seen an example of a beam and brigde problem you should head over to the beam and bridge problems page and try some for yourself.
