Question

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it −135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body?

Answer 1

Answer:

$\beta= 51.65^{\circ}$

$F_r=19.35\ N$

Explanation:

Given:

• force acting to the horizontal right direction, $F_1=10\ lbf$

• force acting -135° to the horizontal, $F_2=25\ lbf$

• force acting 150° to the horizontal, $F_3=5\ lbf$

Now the total components of force in the horizontal direction:

$F_H=F_3\cos 150^{\circ}+F_2\cos(-135^{\circ})+10$

$F_H=5\times cos150^{\circ}+25\times cos(-135^{\circ})+10$

$F_H=-12.0078\ N$ -ve sign means acting in negative x-axis

The total vertical components in the vertical direction:

$F_v=5\times sin150^{\circ}+25\times sin(-135^{\circ})$

$F_v=-15.1776\ N$ -ve sign means acting in negative y-axis

Now the resultant:

$F_r=\sqrt{F_H^2+F_v^2}$

$F_r=\sqrt{(-12.0078)^2+(-15.1776)^2}$

$F_r=19.35\ N$

The angle of the force:

$tan\ \beta=\frac{F_v}{F_H}$

$tan\ \beta=\frac{-15.1776}{-12.0078}$

$\beta= 51.65^{\circ}$