Page 160 - WCS - Electrical

P
BC

θ
Q.Sin
=
α =
tan
AC
Q
θ

BC
P×
AC =
Q
AC

Distance
Speed =

QxBC
Time
P =
Definite
Distance
PxAC

P =
Time
BC

CD

Sin Q× AC P = BC Velocity = P + Q.cos covered direction
θ =
Q a = change in velocity m/sec 2
BD Time
Cos θ =
Q WORKSHOP CALCULATION & SCIENCE - CITS
1

2
2
R = P + Q + 2.P.Q cos θ 2
Momentum = m x v= mass of the body x its velocity Rate of change of momentum = force acting on the body
CD CD  (v −u) 
tan α = = F = m  
OD OB + BD  t 
force = mass x acceleration
Q.Sin θ
=
P + Q.cos θ Applications of Force and Motion
- Understanding the principles of force and motion is essential for various applications, including:
tan α = Q.Sin θ
P + Q.cos θ - Engineering design and analysis
- Transportation systems
2 2 - Sports and athletics
R = P + Q + 2.P.Q cos θ - Aerospace technology
- Machinery and equipment operation
Conclusion
- Force and motion are fundamental concepts that describe the behavior of objects in the physical world.
- Newton’s laws of motion provide a framework for understanding the relationship between force, mass, and
acceleration.
- Applying these principles allows us to analyze and predict the behavior of objects under various conditions.
EXAMPLES
1 If three force of 30, 35, 55 kg respectively act on the same line and in the same direction, find their resultant
force.
Solution: R = P + Q + S
=30+ 35 +55
=120 kg. Ans.
The direction of this resultant force will be in the same direction of all the forces.
2 If a force of 40kg is acting opposite to two forces of 30 kg and 20 kg on the same line, find their resultant force.
Solution: R = P + Q + S
= 30 + 20 – 14
= 10 kg Ans.
This resultant force will act in the direction of 30 kg and 20 kg forces.
3 80 kg force acting on horizontal makes an angle of 300 find its horizontal and vertical components.
Solution: Horizontal components
= F cos Ө
= 80 cos 300
= 80 x √3/2
= 40 x 1.732
69.280 kg
Vertical component = 80 sin 300

= 80 x 1/2
=40 kg Ans.

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CITS : WCS - Electrical - Exercise 13