Page 33 - Physics Form 5 KSSM_Neat

(c) F R Info GALLERY CHAPTER 1
R
The method of calculation will
give a more accurate answer Force and Motion II
compared to the answer obtained
W sin 40° by the method of scaled drawing.
W cos 40° 40°
KEMENTERIAN PENDIDIKAN MALAYSIA
The box is in equilibrium. F = W sin 40 o
R
Resultant force = 0 N = 50 sin 40 o
Forces parallel to the inclined plane are balanced. = 32.14 N
Forces perpendicular to the inclined plane R = W cos 40 o
are balanced. = 50 cos 40 o
= 38.30 N

Example 3
Figure 1.30 shows a poster hanging on the wall of the laboratory
with a string and a nail. The weight of the poster is 12.0 N. T T
(a) Draw the triangle of forces for the weight of the poster and
the tensions in the string. 35° 35°
(b) Calculate the value of T. Love Our Laboratory

Solution
(a) 35°
55°
T
W = Weight of poster
T = Tension in the string W
W = 12 N 70°
Figure 1.30

T
55°
35°
(b) Using the sine rule: Using the cosine rule:
T = 12 W = T + T – 2(T × T × cos 70 )
2
2
2
o
2
2
2
o
sin 55° sin 70° 12 = T + T – 2(T × T × cos 70 )
o
2
T = 12 × sin 55° 144 = T (1 + 1 – 2 cos 70 )
sin 70° 2 144
= 10.46 N T = (1 + 1 – 2 cos 70°)
T = 10.46 N
Formative Practice 1.3
1. State the meaning of forces in equilibrium.
P
2. Figure 1.31 shows a block that is stationary on an inclined
plane when a stopping force, P is applied horizontally.
(a) Sketch and label the weight of the block, W and the
normal reaction from the surface of the plane, R. 30°
(b) Sketch the triangle of forces for P, W and R. Figure 1.31

LS 1.3.3 23