Mechanics 2 Coursework - Ladders

For this model to work for experimental results, we need to make assumptions about some of these values, so that we can calculate a value for µ and also find the value of ? for each position x of the weight. First we assume there to be no friction at the wall, which is quite important because it enables us to find R simply by adding w to mg. This leads to having to define a value for g, which I will take to be 9.81 because that is the average value on Earth (quoted from a Physics Textbook) We also assume there are no errors in the length of the ruler l, enabling us to work out the value of ?. This is less important than my other assumptions, because a small inaccuracy will not make difference to our value.

We can see the way in which the model works below:

First we place the ruler against a smooth wall (the white board) and then let it slide down in a straight line (so that all the forces involved are in the same plane, ensuring a 2d situation). We take 5 readings of each of these heights along the wall, and from these a value of ? can be calculated, enabling us to work out F and therefore µ.

If we assume equilibrium, the ladder's weight w combined with the weight of the person will balance with R (if we resolve forces vertically). By taking moments about the base, we find that S x sin? = (w+mg)cos?. We work out w by weighing the ruler, which turns out to be 72.2g.

From these we can construct a set of equations to find µ:

?=sin-1 z/1 (sin?=O/H)

By N2L Vertically, R=0.0722 + 9.81m

Moments about the base, 1*Ssin? = (0.0722*0.5+9.81mx)cos?

By N2L Horizontally, F = S

F = (0.0722*0.5+9.81mx)cos? / 1*sin?

Friction at limiting equib. µ = F/R

We would not be able to calculate µ unless we assumed there was no friction on the wall, making it a hugely important assumption.

Firstly the experiment was carried out without a weight on top (removing m) to enable us to find a value for µ. This value will not change significantly and applies whenever we have a ruler-on-paper situation. The results are based on an average of 5 readings.

The values we calculated were as follows:

Lower Bound

Average

Upper Bound

74.4

78.46

82

?=sin-1 (78.46/100) = 51.26

µ=F/R, ?=51.26

µ=(0.0722*0.5)cos51.26 / 1 sin51.26*0.0722

This leaves us with a value of roughly 0.4 for µ. We can use this in all of our future calculations.

Chapter 2: Predictions and Experiment

In this chapter some basic predictions about the actions of the ruler can be made, based on the value of µ calculated in the previous chapter and also on the assumptions that were made. We can firstly predict R for all cases, because it does not have an x in its equation.

By N2L, R=0.0722 + 9.81x0.2 = 2.03

The values of x I will measure are not uniform, because this shows that x does not have to be uniform to conform to the predictions made. I used values of 90cm, 75cm, 60cm, 40cm and 20cm. To predict the value of ? at which the ruler will slip, I can put these values for x into the previous equations:

x=0.2, R= 2.03

Moments about base, 1*Ssin? = (0.0722*0.5+9.81*0.2*0.2)cos?

S = F = µR = 0.4*2.03 = 0.812

sin?/cos? = (0.0722*0.5+9.81*0.2*0.2)/1*0.812 = 0.528

tan?=1/0.528 ?=62.17

The value for z from this value of ? comes to 0.884. The others are as follows:

x

?

z

0.2

62.17

0.884

0.4

44.68

0.703

0.6

33.79

0.556

0.75

28.30

0.474

0.9

24.26

Alex Hayton Mechanics 2 Coursework - Page 1 of 1

Ladders