Numerical Problem No. 4.3 (Turning Effect of Forces) Physics SSC - I (Class 9)

Find the magnitude and direction of a force, if its x-component is 12 N and y- component is 5 N. Ans. (13 N making 22.6° with x-axis)

Given:  

Horizontal Component of Force  = F_x =   12 N

Vertical Component of Force  = F_y =   5 N 

To Find:

Resultant Force = F = ? N

Direction (Angle) of Force = 𝜭 = ?

Solution:  

The formula for the resultant force F is

F = `\sqrt {{Fa}^2 + {Fy}^2}`

by putting values

F = `\sqrt {{12N}^2 + {5N}^2}`

F = `\sqrt {144N^2 + 25N^2}`

F = `\sqrt {169 N^2}`

F = 13 N   

Thus the magnitude of the resultant force is 13 N

Now to find the direction of the resultant force we have 

Tan 𝜭 = `\frac {Fx}{Fy}`

or 

𝜭 = Tan⁻¹( `\frac {Fy}{Fx}` )

by putting values

𝜭 = Tan⁻¹( `\frac {13 N}{12 N}` )

𝜭 = Tan⁻¹( 0.417 )

𝜭 = 47.291°

thus the direction of the resultant force is 47.291° with the x-axis.

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