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Mathematics

 

 

Improving the Theoretical Model

The ratchet socket travels a shortest distance d is than the theoretical model predicts this could be due to the socket travelling slower than calculated. This is most likely due to friction between the object and the slide. Initial assumptions must be reconsidered, there is a resistive force going against the object as it travels down the slide. The graph, below, shows how differences vary between theoretical and experimental models over the three values of q .

When the object is on a slight slope, q = 24 the resistive forces are greatest, with almost 20cm difference when the ratchet socket is placed 1 metre from the edge of the slide. Resistance at q = 30 appears to be smaller than both greater and lesser values of q . This is possibly where resistive forces are minimal between the socket ratchet and the slide compared to q = 24; the ratchet socket is projected more horizontally than q = 35.

q = 30 and q = 35 have similar values for d. At q = 30 friction will be greater than q = 35, but at q = 35, where the slope is steeper, the ratchet socket is propelled more vertically downwards than at smaller values of q . The similar values for d between q = 30 and q = 35 allows for the experimental model to catch up with the theoretical model, therefore differences are smaller between the two.

 

To find the resistive force R, fundimental laws must be applied and established formulae can be used with them.

F = ma F = mgsin q F can now be written as a known value

mgsin q R = ma Where R is the resistance to be found

a = gsin q R/m m is divided out to leave a

To find R/m the median values of d will be used from the experimental model. As d = vtcos q when the object starts at rest, t =. The velocity of the object must be calculated firstly, the position of the object horizontally will be used to calculate v. To find R/m the new formulae must be found by working backwards from where the ratchet socket lands on the floor back to the start, on the slide.

h = vtsin q gt2 t =and h = 0.71

h = dtan q g replaces t as both are equal to each other, eliminates v

h = dtan q Each term is squared individually, so terms can be seperated

h dtan q = dtan q is subtracted

2v2cos2 q = 2v2cos2 q multiplied out of the divisor

v2 = 2cos2 q divided out to leave v2

v2 = Simplified

v2 = u2 + 2as s = l in the context of the model

= u2 + 2al As v2 has now been found, terms in u2 + 2as are known

= 0 + 2(gsin q R/m)l

= gsin q R/m

R/m = gsin q where v2 =

 

R/m could be better than theoretical model as it takes into consideration the resistance against the ratchet socket, probably due to friction.

A2 = l, (F2/100) = d, H2 = h, 9.8 = g

R/m

=(9.8*SIN(RADIANS(24)))-((9.8*POWER((F2/100),2))/
(2*POWER(COS(RADIANS(24)),2)*(H2-(F2/100)*TAN(RADIANS(24)))))/(2*A2)

As the experimental tests were carried out using centimetres, the median value of d in F2 had to be converted to metres before being calculated. This does not effect the outcome in any way as "F2/100" is enclosed in brackets.

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