Matemática, espíritu y arte: septiembre 2015

Let's see an exercise.
A 40 kg box is dragged 30 m on a horizontal floor, applying a Fp = 100 N exerted by a person. Such force acts doing a 60º angle. The floor exerts a friction force Fr = 20 N. Calculate the work done for each one of these forces Fp, Fr, the weight mg and the normal. Calculate also the net work done on the box.

Solution: There are four forces acting on the box, Fp, Fr, the weight mg and the normal.
The work done by mg and the normal N is zero, because they are perpendicular to displacement(=90º for them).
The work done by Fp is:
Wp = Fpxcos (using x in place of d) = (100 N)(30 m)cos60º = 1500 J.
The work done by the friction force Fr is:
Wr = Frxcos180º = (20 N)(30 m)(-1) = -600 J.
The angle between Fr and displacement is 180º because they point in opposite directions.
The net work can be calculated in two equivalents ways:

As the algebraic sum of the work done by each force:
W_NET = 1500 J +(- 600 J) = 900 J.
Finding first the net force on the object along the displacement:
F_(NET)x= Fpcos - Fr
and then doing
W_NET = F_(NET)x = (Fpcos - Fr)x
= (100 Ncos60º - 20 N)(30 m) = 900 J.

Ohm's Law,
Example Two
Determine the total resistance and potential difference between points a and b in the circuit shown. Also determine the current in, and the potential difference across, each resistor. The input current is 60 A.

a) Total resistance between a and b.

First, calculate the 12, 6, and 4 ohm parallel combination R_//1
1/R_//1 = 1/12 + 1/6 + 1/4 = 1/2. Then R_//1 = 2 ohm.

The upper branch is 3 ohms in series with R//1, 3 ohm + 2 ohm = 5 ohm.

This 5 ohm resistor is in parallel with the 20 ohm resistor, equivalent to a single resistor R_//2 = 5 ohm//20 ohm:
1/R_//2 = 1/5 + 1/20 = ¼. Then R_//2 = 4.
The total resistance between a and b is 4 ohm + 5 ohm = 9 ohms. The voltage drop or potential difference between points a and b is 60Ax9ohm = 540 V.

b) To obtain the individual currents and voltages in each resistor we can proceed as follows (there is no a unique method):

The voltage across R_//2 is the difference between the total 540 voltage and the voltage drop in the 5 ohm resistor. This drop is 60Ax5ohm = 300 V. So the potential difference across R_//2 is 540 V – 300 V = 240 V.
Now we can calculate the currents and voltages in the R_//2combination:
In the 20 ohm resistor: current= 240 V/20 ohm = 12 A, voltage = 240 V
The upper branch has then a current of 60 A - 12 A = 48 A, or alternatively 240 V/5 ohm = 48A.
The 3 ohm resistor has a voltage of 48 Ax3 ohm = 144 V.
The R_//1 combination presents a voltage of 48 AxR_//1 = 48 Ax2 ohm = 96 V, or alternatively 240 V - 144 V = 96 V.
The currents in the 12 ohm, 6 ohm and 4 ohm are respectively 96/12 A=8 A, 96/6=16 A and 96/4=24 A