ALLIANZE UNIVERSITY COLLEGE OF MEDICAL SCIENCES FOUNDATION OF MEDICAL STUDIES 2011
PHYSICS LABORATORY REPORT
EXPERIMENT 1: STATIC AND DYNAMIC
NAME
: NURUL HASANAH BINTI NOR IKHSAN
MATRIX NO. : CPM 33/11C GROUP
:A
SEMESTER : III DATE
: 26.04.2011
TITLE Static and dynamic OBJECTIVE To determine the coefficient of a) Static friction b) Kinetic friction
THEORY
Figure 1 Forces of friction is when an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion. This is due to the interactions between the object and its environment. Friction that prevents motion from occurring is called static friction, fs while the one that impedes a motion in progress is called kinetic friction, fk. Referring to Figure 1, if force, F is increased, frictional force, f also increases and the object will remain at rest (static). However, for a certain value of F, the object starts to move. The frictional force at this stage is known as the limiting static friction force, fs which is the maximum value of f. fs = µsR fs = µsmg where µs = coefficient of static friction fs =static frictional force R = normal reaction
When the object is in motion, the frictional force is known as kinetic friction, fk . The kinetic frictional force is less than the static frictional force. That explains why it is difficult to move an object which is initially at rest, but once it is set in motion, less force is needed to maintain the motion. Fk = µlR where µk = coefficient of static friction fk =static frictional force R = normal reaction Since fk = fs , therefore µs > µk
APPARATUS a) b) c) d)
A wooden block Two sets of slotted mass with hooks A set of pulley with clamp Plasticine
PROCEDURE
Figure 2 a) The mass of the wooden block, mb is weighed. b) The apparatus is set up as in Figure 2. The string from the block must be tied up horizontally to the pulley. c) The slotted mass is added onto the hook gradually until the wooden block began to slip. The total mass, mh is recorded. The step is repeated three times to get the average value of mh. d) Different masses, mr are added onto the wooden block and step (c) is repeated. e) Step (d) is repeated for at five different values of mr. f) A graph of fs against R is plotted where fs = mhg and R = (mr + mh)g. The coefficient of static friction, µs is determined. g) Step (c) is repeated with a little push exerted to the wooden block every time each mass is added. The mass, mh is recorded when the block moved slowly and steadily along the plywood. h) Different masses, mr are added onto the wooden block and step (g) is repeated. i) Step (h) is repeated for five different values of mr. j) A graph of fk against R is plotted where fk = mhg and R = (mr + mh)g. The coefficient of kinetic friction, µk is determined from the graph. k) The results are recorded and tabulated in the table.
RESULT
PART A Mass of wooden block, mb Gravity
: 0.234 kg : 9.8 ms-2
mb + mr (kg) 0.2340 0.2540 0.2740 0.2940 0.3140
mh (kg) mh3
mh1
mh2
0.0510 0.0560 0.0560 0.0710 0.0710
0.0510 0.0560 0.0660 0.0610 0.0710
0.0460 0.0560 0.0610 0.0610 0.0710
f s (N)
R (N)
0.4802 0.5488 0.5978 0.6301 0.6958
2.2932 2.4892 2.6852 2.8812 3.0772
mh ave 0.0490 0.0560 0.0610 0.0643 0.0710
Graph of fs(N) against R(N) 0.8 0.7 0.6 0.5
fs(N) 0.4 0.3 0.2 0.1 0 2.2932
2.4892
2.6852
R(N)
2.8812
3.0772
PART B Mass of wooden block, mb Gravity
: 0.234 kg : 9.8 ms-2
mb + mr (kg) 0.2340 0.2540 0.2740 0.2940 0.3140
mh (kg) mh3
mh1
mh2
0.0410 0.0460 0.0510 0.0510 0.0560
0.0410 0.0460 0.0510 0.0560 0.0560
0.0410 0.0460 0.0460 0.0510 0.0560
f k (N)
R (N)
0.4018 0.4508 0.4831 0.5155 0.5488
2.2932 2.4892 2.6852 2.8812 3.0772
mh ave 0.0410 0.0460 0.0493 0.0526 0.0560
Graph of fk(N) against R(N) 0.6 0.5 0.4
fk(N) 0.3 0.2 0.1 0 2.2932
2.4892
2.6852
R(N)
2.8812
3.0772
DISCUSSION In both experiments, the coefficient of static and kinetic friction µs and µ k. can be determined from the gradient of the graph as each graph is and fk against R respectively. Therefore, in part A, the coefficient of static friction, µs is : fs = µs R µs = fs/R = y2 – y1 x2 – x1 = (0.70 - 0.48) (3.08 - 2.29) = 0.22/0.79 = 0.28 In part B, the coefficient of kinetic friction, µ k. is : f k = µk R µk = fk/R = y2 – y1 x2 – x1 = (0.55 - 0.40) (3.08 - 2.29) = 0.15/0.79 = 0.19 Static friction is a force that acts to keep an object from moving. At rest, the value of force exerted (F) is equal to the value of static friction force (fs). Therefore, when F increases, fs also increase. Theoretically, the value of µs is greater than µk as it needs more force to move an object from rest. But, when the object starts to move, less force is needed to maintain the motion. Due to this, in the first experiment of static friction, no external force (push) is applied on the block. While on the second experiment of kinetic friction, an external force (push) is applied every time the masses are added to enable the block to move. This is because kinetic friction is the force that acts when the object is in motion. This experiment can be said successful because the value of µs > µk. in which µs is 0.28 and µk is 0.19.
CONCLUSION The coefficient of static friction, µs is bigger than kinetic friction, µk. Hence, more force is needed to move an object at rest but less force is needed to maintain the motion of a moving object.
PRECAUTION a) The slotted mass must be added gently and carefully to avoid impulsive force acting on the object which can ruin the actual result of the experiment. b) The wooden block must be weighed together with the tied string and plasticine to get an accurate mass of the wooden block. c) The force applied on to the object must be constantly exerted to prevent any inaccurate results.
REFERRENCES
Laboratory manual, Static and Dynamic. Chapter 2: Dynamic and Rotational Motion by Pn. Arihasliza Ariffin.
