4.3newtons 1st Law and 2nd Law
4.3newtons 1st Law and 2nd Law
4.3newtons 1st Law and 2nd Law
For Objects at Rest: If an object is at rest, i.e. there, velocity and acceleration are zero.
The object will remain at rest until an external force is applied.
For Objects in Motion: If an object is in a state of uniform motion, i.e. it has an initial
velocity. The object will remain in a state of uniform motion until an external force is
applied.
The net force on an object is defined as the vector sum of all external forces exerted on the
object.
External forces come from the object’s environment.
Internal forces, which are forces that occur between objects within a system, cancel each other
out in terms of the motion of the system as a whole. Therefore, internal forces do not contribute
to changes in the object’s state of motion, and Newton’s first law focuses on the effects of
external forces that can cause changes in motion.
T₁
√3 =T₂ √2
2 2
T₂=
√3 T ₁
2
Resolution of forces along vertical direction gives;
T₁ sin 30⁰ + T₂ sin 45⁰= 100N
1 3
√
√2
T 1 × + T 1 × =100 N
2 2 2
T1(1+√3)=200N
T1=73.2N
T2=89.65N
2. Two forces of 15 N and 25 N are applied to an object in the same direction. What is the
net force acting on the object?
Solution:
When forces are applied in the same direction, you can find the net force by simply adding
them.
F net= 15N+25N=40N
3. Two masses of 𝑚1 = 2𝑘𝑔 and 𝑚2 = 8𝑘𝑔 are connected by a mass less string. They are
Supported on a frictionless horizontal surface. A horizontal force of 𝐹 = 40𝑁 is applied to
the mass, 𝑚2, as shown. Calculate the tension in the string between the two masse.
Solution:
In the y-direction, the net force in each mass is zero, i.e., the weights and the normal forces are
equal but opposite, and hence. ∑ 𝐹𝑦 = 0
On the other hand, the horizontal forces acting on the masses are:
Mass, m1: ∑ 𝐹𝑥 = 𝑇 = 𝑚1𝑎𝑥….. (a)
Mass, m2: ∑ 𝐹𝑥 = 𝐹 – 𝑇 = 𝑚2𝑎𝑥…… (b)
Substituting Eq. (a) into (b), we get ∑ 𝐹𝑥 = 𝐹 – 𝑚1𝑎𝑥 = 𝑚2𝑎𝑥,
So that the acceleration of the masses is
Fx 40 N
ax= = =4 m/ s ²
m 1+ m2 10 Kg
Then, substituting the value of the calculated acceleration into Eq. (a), we find the tension, T, in
the string to be:
𝑇 = 𝑚1𝑎𝑥 = (2𝑘𝑔)(4𝑚/s²)=8N