Activity No. 3
COEFFICIENT OF FRICTION
OBJECTIVE:
To measure the coefficient of friction between two surfaces.
APPARATUS:
1
Block of wood
1
Dynamic cart
1
Triple beam balance
1
Friction board with pulley
3
2
1
1
100 g mass
200 g mass
50 g mass
10 g mass
1
1
1
1
Cord
Meter stick
can
Stopwatch
INTRODUCTION:
The experiment measures the coefficient of friction between two wood surfaces. The
set-up is shown below:
Figure 3.1
To meet the objective, the forces acting on the system must be identified – this must
be done for the cart, the block and the can.
After this, Newton’s Laws are applied to establish the equations of motion. Note that with
the system viewed in this perspective, the cart and the block will move to the right and
the can will move downward, this accounts for the assignment of the signs of the
acceleration later.
The forces acting on the cart are tension, normal force and weight,
as shown in the free body diagram (FBD).
By applying Newton’s Laws, the corresponding equations are
∑ 𝐹𝑥 = 𝑇1 = 𝑚𝑐𝑎𝑟𝑡 𝑎
(1)
Laboratory Manual in ES 15A – Physics for Engineers
Figure 4 2
10
∑ 𝐹𝑦 = 𝜂𝑐𝑎𝑟𝑡 − 𝑤𝑐𝑎𝑟𝑡 = 0
(2)
The forces acting on the block are tension from two strings, normal force, weight and
friction. The forces are shown in the FBD.
By applying Newton’s Laws, the corresponding equations are
∑ 𝐹𝑥 = 𝑇2 − 𝑇1 − 𝑓 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑎
(3)
∑ 𝐹𝑦 = 𝜂𝑏𝑙𝑜𝑐𝑘 − 𝑤𝑏𝑙𝑜𝑐𝑘 = 0
(4)
The forces acting on the can are tension and weight.
By applying Newton’s Laws, the corresponding equations
are
∑ 𝐹𝑦 = 𝑇2 − 𝑤𝑐𝑎𝑛 = −𝑚𝑐𝑎𝑛 𝑎
Figure 4 3
(5)
From equation (1): 𝑇1 = 𝑚𝑐𝑎𝑟𝑡 𝑎
From equation (5): 𝑇2 = 𝑤𝑐𝑎𝑛 − 𝑚𝑐𝑎𝑛 𝑎 = 𝑚𝑐𝑎𝑛 𝑔 − 𝑚𝑐𝑎𝑛 𝑎
Substituting the values of T1 and T2 to equation (3):
𝑇2 − 𝑇1 − 𝑓 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑎
𝑚𝑐𝑎𝑛 𝑔 − 𝑚𝑐𝑎𝑛 𝑎 − 𝑚𝑐𝑎𝑟𝑡 𝑎 − 𝑓 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑎
Figure 4 4
Note that the force of friction is given by
𝑓 = 𝜇𝜂𝑏𝑙𝑜𝑐𝑘 = 𝜇𝑤𝑏𝑙𝑜𝑐𝑘 = 𝜇𝑚𝑏𝑙𝑜𝑐𝑙 𝑔
So, equation (3) becomes,
𝑚𝑐𝑎𝑛 𝑔 − 𝑚𝑐𝑎𝑛 𝑎 − 𝑚𝑐𝑎𝑟𝑡 𝑎 − 𝜇𝑚𝑏𝑙𝑜𝑐𝑘 𝑔 = 𝑚𝑏𝑙𝑜𝑐𝑘 𝑎
After some algebraic manipulation,
(𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛 + 𝑚𝑐𝑎𝑟𝑡 )𝑎
= 𝑚𝑐𝑎𝑛 − 𝜇𝑚𝑏𝑙𝑜𝑐𝑘
𝑔
The acceleration a, when x is constant, is given by 𝑎 =
2𝑥
𝑡2
Substituting the value of the acceleration to equation (3):
(𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛 + 𝑚𝑐𝑎𝑟𝑡 )𝑎
= 𝑚𝑐𝑎𝑛 − 𝜇𝑚𝑏𝑙𝑜𝑐𝑘
𝑔
(𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛 + 𝑚𝑐𝑎𝑟𝑡 )(2𝑥)
= 𝑚𝑐𝑎𝑛 − 𝜇𝑚𝑏𝑙𝑜𝑐𝑘
𝑔𝑡 2
Let
Laboratory Manual in ES 15A – Physics for Engineers
11
𝑘=
(𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛 + 𝑚𝑐𝑎𝑟𝑡 )(2𝑥)
𝑔
𝑇=
1
𝑡2
Then, equation (3) becomes
𝑘𝑇 = −𝜇𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛
𝜇
𝑚𝑐𝑎𝑛
𝑇 = − 𝑚𝑏𝑙𝑜𝑐𝑘 +
𝑘
𝑘
From this, the coefficient of friction can be determined by solving for the slope of the
line that best fit the graph of mb vs T. The slope is given by
𝜇
𝑚=−
𝑘
Procedure:
1. Get the mass of the cart, block and wood using the triple beam balance. Convert your
2.
3.
4.
5.
6.
7.
masses in kg unit. Record this in Table 4.1 #1 a, b, and c respectively.
Choose the distance 𝑥 you like to set in the experiment using meter stick. Convert
your value of 𝑥 in meter. Put your value in Table 4.1 #2. This distance is the length
of the block from the stopper.
Put 360 g of slotted masses on the cart, 200 g of slotted masses on the block, and
200 g of slotted masses on the can. This will serve as their initial loads.
Using the initial loads and distance, record the time it takes for the block to reach the
stopper. Do it for 5 trials. Put your recorded time for 5 trials in Table 4.1 #3 under
Trial 1, 2, 3, 4 and 5. This will be your data for 200-g mass of block.
For 250-g mass of block data, take 50 g of mass from the cart and place it on the
block. Then record the time it takes for the block to reach the stopper for 5 trials. The
200-g load of can is fixed. Only the loads on the cart and block will vary.
Repeat procedure #5 for 300 g, 350 g, 400 g, 450 g, 500 g, and 550 g of load of
block for 5 trials each. Record these in Table 4.1.
Compute for the tave, δt, T, δT, T+δT, and T-δT to all data. Necessary equations are
given below.
𝑡𝑎𝑣𝑒 =
Laboratory Manual in ES 15A – Physics for Engineers
𝑡1 + 𝑡2 + 𝑡3 + 𝑡4 + 𝑡5
5
12
(𝒕𝒂𝒗𝒆 − 𝒕𝟏 )𝟐 + (𝒕𝒂𝒗𝒆 − 𝒕𝟐 )𝟐 + (𝒕𝒂𝒗𝒆 − 𝒕𝟑 )𝟐 + (𝒕 − 𝒂𝒗𝒆 − 𝒕𝟒 )𝟐 + (𝒕𝒂𝒗𝒆 − 𝒕𝟓 )𝟐
√
𝜹𝒕 =
𝟓
𝑇=
𝛿𝑇 =
1
(𝑡𝑎𝑣𝑒 )2
2𝛿𝑡
(𝑡𝑎𝑣𝑒 )3
T + δT and T - δT are just the sum and the difference of T and δT respectively.
To solve for coefficient of kinetic friction k:
𝑘=
(𝑚𝑏𝑙𝑜𝑐𝑘 + 𝑚𝑐𝑎𝑛 + 𝑚𝑐𝑎𝑟𝑡 )(2𝑥)
𝑔
The slope m can be obtained through linear regression given below,
𝑚=
(𝑁 ∑ 𝑋𝑌) − (∑ 𝑋 ∑ 𝑌)
(𝑁 ∑ 𝑋 2 ) − (∑ 𝑋)2
where
and
N = the number of data points = 5
X = masses on the block, mblock
Y = values of T.
Similarly, the maximum slope mmax and minimum slope mmin is determined through linear
regression above.
For mmax:
X = masses on the block, mblock
and Y = values of T + δT.
For mmin:
and
X = masses on the block, mblock
Y = values of T - δT.
As discussed earlier, the coefficient of friction is 𝝁 = −𝒎𝒌 while the uncertainty of the
𝒌
coefficient of friction, Δ𝜇 is 𝚫𝝁 = (− 𝟐)(𝒎𝒎𝒂𝒙 − 𝒎𝒎𝒊𝒏 ) .
Laboratory Manual in ES 15A – Physics for Engineers
13
GUIDE QUESTIONS
Table 4.1
1. What are the masses of the following:
a. Dynamic cart (mcart)
___0.677_______ kg
b. Wooden block (mblock)
____0.404_________ kg
c. Can (mcan)
_____0.253________ kg
2. distance x of the block from the stopper
______1.0_________ m
3. How much time does it take for the block to travel the distance x for 5 trials each
when it is loaded with 200g, 250g, 300g, 350g, 400g, 450g, 500g, and 550g?
TRIAL (s)
mblock
tave
δt
T
δT
T+δT
T-δT
(s)
(s)
(1/s2)
(1/s2)
(1/s2)
(1/s2)
(g)
t1
200
1.17
1.30 1.42 1.34
1.19
1.28
0.09
0.61
0.09
0.70
0.52
250
1.41
1.50 1.48 1.44
1.49
1.46
0.03
0.47
0.02
0.49
0.45
300
1.59
1.72 1.66 1.64
1.66
1.65
0.04
0.37
0.02
0.39
0.35
350
1.74
1.74 1.72 1.71
1.68
1.72
0.02
0.34
0.01
0.35
0.33
400
1.82
1.78 1.85 1.74
1.89
1.82
0.05
0.30
0.02
0.32
0.28
450
1.98
1.95 1.93 2.04
1.90
1.96
0.05.
0.26
0.01
0.27
0.25
500
2.06 2.12 1.98 1.99 2.08 2.05
0.05
0.24
0.01
0.25
0.23
550
2.46 2.13 2.02 2.34
0.16
0.20
0.03
0.23
0.17
t2
t3
t4
t5
2.19
2.23
4. What is the value of k?
0.27 J
5. What is the slope m of the graph of mblock vs T?
6. What is the maximum slope mmax?
7. What is the minimum slope mmin?
8. How much is the coefficient of friction, μ?
Laboratory Manual in ES 15A – Physics for Engineers
1.27 J
1.36 J
1.13 kg
-0.34
14
9. How much is the uncertainty of the coefficient of
friction, Δμ?
-0.35
CONCLUSION:
Therefore, We determined the value of coefficient of friction between two surfaces which is -
0.35.
Laboratory Manual in ES 15A – Physics for Engineers
15