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Rectilinear Motion

If a particle of mass m accelerates in the x direction, the force in x direction is given by

(according to Newton’s 2nd law of motion).

 

Ʃ F x = m a x [ scalar equation ]

Ʃ F y =0, Ʃ F z =0

 

In general case if the particle of m, has components ax, ay and a z in three directions, we have

Ʃ F x = m a x  

Ʃ F y = m a y

Ʃ F z = m a z

UNIT OF FORCE : S.I. unit of force is Newton (N).

l N = l kg x l m/s2 = 1 kg m/s2.

 

 

 

 

SOLVED EXAMPLES

 

Example 5.1. A force acts on a body of mass 15 kg and produces an acceleration of 6 m/s2 in the direction of force. Determine_ tlze magnitude of the force.

 

Solution. Given                    m =· 15 kg, a = 6 m / s2

Force F= ma

=15 x 6kg m/s2.

= 90 N. Ans

 

Example 5.2. A force of 1000 N acts on a body having mass of 8 kg for 10 seconds. If the initial velocity of the body is 10 m/s, determine (i) acceleration produced in the direction of force, and (ii) distance moved by the body in 10 seconds.

 

Solution.                        F= ma

1000 = 8 x a

a = 1000 / 8 = 125 m / s

v0 =10 m /s ,t =10 + 125 x10 = 1260 m / s2

v = v0 +at = 10 + 125 x 10 = 1260 m/s

v2 = v0 +2 x a x s

(1260)2 =(10)2 +2 x 125 x S

S = (1260)2 – (10)2 / 2 x 125 = 6350 m.

 

Example 5.3. The weight of a body on earth is 981 N. If the acceleration due to gravity on earth is 9.81 m/s2, what will be weight of the body on (i) moon (g moon = 1.7 m/s2), (ii) Sun (g sun = 275 m/s2).

 

Solution. W = mg

981 = m x 9.81

m = 981 / 9.81 =100 kg

 

(i) The mass remains constant.

W at moon= m x g moon

= 100 x 1.7 = 170 N. Ans.

 

(ii) Weight at sun

W = m x g sun

=100 x 275 27500 N. Ans.

 

Example 5.4. A force of 20 N acts on a body having mass of 30 kg for 90 seconds. If the initial velocity of the body is 2 m/s, determine the final velocity of the body when (i) the force acts in the direction of motion ; (ii) the force acts in the opposite direction of motion.

 

Solution :                20 = 30 x a

a = 2/3 m/s2

v = v0 + at = 2 +2 /3 x 90 = 62 m/s .Ans.

 

 

(ii)                                            force = -20

a = -20 /30 = -2 / 3

v = v0  +at = 2 / 3 x 90 = – 58 m / s . (negative sign shown

the body will move in the direction of the force which is opposite of case (i))

 

 

Example 5.5. A body of mass 30 kg falls from rest on the ground from a height of 20 m. If the resistance of the ground against penetration is 5000 N, find tl1e depth of penetration.

 

Solution.              v0 = v02 + 2g x 20

= 0 + 2 x 9.81 x 20

= 19.81 m/s.

Weight of the body,    W = mg = 30 x 9.81

= 294.3 N.

Net upward force = 5000 – 294.3

= 4705.7 N  ; 4705.7 / 30 = a (retardation)

v2 = v02 – 2aS

0 = (19.81)2 – 2 x 4705.7 /30 x S

S = 1.251 m. depth of penetration .Ans.