Using The Principles of Simple Machines To Teach Algebra II

 As I've written before, teaching basic physics and math can be challenging especially if not done in concrete ways. I learned this firsthand in my first two years in Peace Corps teaching algebra to 2nd and 3rd formers (the equivalent of U.S. high school freshmen and sophomores) in Barbados.

For example, on first called upon to teach groups and group theory to 2nd formers, I found the use of 'clock groups' of particular benefit, as noted in an early  (2010) blog post:

Brane Space: Looking at Groups

However, teaching the equivalent of algebra II proved more challenging, until I realized that the incorporation of practical devices could help immensely. Thus, I launched the beginning use of simple machines and then applying aspects of them to solving multiple problems.

I found that an inclined plane machine such as shown below, can be used to show how to solve inclined plane problems:

This then can lead to preparation of applicable force diagrams:

                                                                     

We note in Fig. 2 the force of gravity (W)  into components tangent and normal to the plane.  So long as the block does not move, then:

N =  mg cos q      F = mg sin q

 We also see F increases as the inclination of the block increases. Hence, the inclination angle q cannot exceed a certain value (call it  q_max ).  When the block is at the point of slipping we have:  

      mg sin q_max  =  m_s N   =      m_s mg cos q  

The coefficient of static friction is then obtained by dividing the far left side of the equation by the right side to get;

      mg sin q_max /mg cos q_max   =   tan  q_max  =  m_s

   Because there is no acceleration perpendicular to the plane: 

N - mg cos q_max   = 0

A typical problem might be:

     For a particular inclined plane tan q = 0.2875. Given a block of mass 0.3 kg resting midway on the incline, find the coefficient of static friction m_s. 

 Solution:  

For the block to remain in equilibrium, we need the tangential and normal forces (F  and N)   to balance: 

mg sin q  =   m_s mg cos q.   

Divide both sides by mg cos q:

 sin q/cos q  =   m_s

But:    sin q/cos q =   tan q = 0.2875

So:  m_s  = 0.2875

The inclined plane can also be treated as a simple machine.

The theoretical M.A. or "ideal mechanical advantage" is defined: M.A. = mg/F = W/F, while the actual mechanical advantage is defined: M.A.(ac) = L/h. Meanwhile work is defined as the force x distance moved, so the work done in pulling the block up the plane is: w = F' x L, and this is also equal to the gain in potential energy, W x h. The efficiency of any given machine is defined:

Efficiency = work output/ work input or Wh / F'L = mgh/ F'L.

Figure 2 illustrates the condition in which friction isn't omitted and we let W be the weight of the object resting on the inclined plane. This object, say block, tends to slide down the plane under the action of force F = mg sin Θ. But, if now a force F' is applied to pull the object up then we need the magnitude of F' >   F.  I.e. to overcome the force of friction, F  between the object and the plane. Thus: F' = F + F

We know: F = mg sin Θ and F = m_s N = m_s mg cos Θ, so that:

F' = mg sin Θ + m_s mg cos Θ, or F' = mg(sin Θ + m_s cos Θ).

Other machines can also be used starting with a simple pulley system:

A basic pulley system, is depicted in Fig. 3(a) alongside a modification in Fig. 3(b). The pulley shown in 3(a) is a single movable pulley, in contrast to the Atwood machine which is a single fixed pulley. In operating such a pulley, say to lift a weight w, the force applied (F) must move twice as far as the weight w = mg. The mechanical advantage (assuming no friction) is s/d which is the displacement of the applied force, how much it moves, divided by the distance the weight is moved. Since for Fig. 3(a) if the weight w is moved 1' then the force F is moved 2'. Thus, s = 2' and d = 1' so: w/F = s/d or F = wd/s = ½ w.

In Figure 4 a variety of different pulley systems is depicted including a variant called the "wheel and axle" (A) and one of grouped pulleys (C) and multiplied strings(B). The wheel and axle is of particular interest in that it makes use of two different radii, an inner small one, r and a larger outer one R. If the depicted wheel (Fig. 4(A)) moves through one complete revolution, the distance the force will move is just d = 2πr. Meanwhile, the distance the force moves will be s = 2πR. If we take the mechanical advantage:

M.A. = s/d = (2πR)/ (2πr) = R/r, then:

mg/ F = w/F = R/r and so: F = (r/R)w which is the law of the wheel and axle.

Lastly, we come to perhaps the most famous machine of all, the lever. A basic depiction of a workable lever is shown in Figure 5.

  Basically a load L is placed at one end which we wish to lift by applying a force F at the other end. Let the load be a distance a from the pivot, and the applied force acts at a distance, b. Then:

Force x distance from axis = load (mg) x distance from axis or:

F x b = L x a or F = (a/b) L = (a/b) mg.

This is called the "law of the lever". It helps to illustrate using a simple problem how it works:

Sample problem:

A 50 kg concrete block has to be moved from the ground to a wheelbarrow and a workman is provided with a board 5 m in length. If the workman pivots the block at 3.5 m from one end and lifts from the other (assume g = 10 m/s ²) What applied force is needed to lift the block? What is the work done?

We have the effort distance, a = 5.0 m - 3.5 m = 1.5 m, and the load is:

w = mg = 50 Kg (10 m/s ²)= 500 N, with load distance a = 3.5m. Then, since:

F x b = L x a, we have:

F = (a/b) w and F = (1.5 m/ 3.5m) 500 N = 214 N.

The work done is Fs = (mg)d but (d/s) = (a/b) so Fs = (a/b) mg x 1.5 m = 321 J.

Algebra II Word Problems based on simple machines:

1) Show, with reasons, that the pulley system shown in Fig. 3(b) has a mechanical advantage of 2. If the weight w is lifted 1m, how much distance must the force F cover?

2) An inclined plane has an angle of Θ = π/6. The coefficient of sliding friction u = 0.3.How much force, parallel to the incline, is needed to push a 100 N object up the incline at constant speed?

3) The work done in moving a 50 kg mass up an incline 1 m high and 5m long is 640 Joules. What is the frictional force in N?  Compare the ideal and the actual mechanical advantage.

4) In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.

5) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 m/s ², find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done?

6)A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.

7) In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.

See Also:

• Kids Hate Algebra II? WHY?