Question

AB is a rigid rod of 6m long, at the ends of A and B two like parallel forces of magnitude P=5N and Q = 10N are acting. Find the magnitude, direction and point of application of the resultant (diagram is mandatory)​

Answer 1

Answer:

Two like parallel forces 20 N and 30 N act at the ends A and B of a rod 1.5 m long. The resultant of the forces will act at the point:

Answer 2

Given :

• Length of rod AB = 6m
• Magnitude of force A (P) = 5N
• Magnitude of force B (Q) = 10N

To Find :

• Magnitude of resultant force(R)
• Direction of resultant force.
• Point of application of resultant force.

Formulas Applied :

Magnitude : The magnitude of the resultant force is the sum of Force's of A and B. hence,

${\sf Magnitude_{(R)} =Magnitude_{(A)}+ Magnitude_{(B)}}$

$\boxed{\sf R = P + Q}$

Direction : The resultant acts in a direction same as that of 'P' and 'Q'.

Point of application : The resultant acts at a point 'C' in between A and B.Such that,

$\boxed{ \sf P \times AC = Q\times BC}$

Solution :

${\longrightarrow \sf Magnitude = R = P + Q}\\\\{\longrightarrow R = 5+10}\\\\\longrightarrow \boxed{\sf R=15N}$

Point of Application :

${\longrightarrow \sf P \times AC = Q \times BC}\\\\{\longrightarrow \sf 5 \times x = 10(6-x)}\\\\{\longrightarrow \sf 5x = 60-10x }\\\\{\longrightarrow \sf 5x+10x = 60}\\\\{\longrightarrow \sf 15x = 60}\\\\{\longrightarrow \sf x = \dfrac{60}{15}}\\\\{\longrightarrow \sf x = 4m}$

The resultant acts at a point 'C', 4m from 'A'.