Question

30N and 46N forces are perpendicular to each other and pull a single object what is the third force that must act on the object if the result is 0

Answer 1

Let $F_1=30\ N$ pulls the object along the horizontal and $F_2=46\ N$ pulls the object along the vertical. Let's find the magnitude and direction of the third force. Let’s write the sum of projections of the forces on axis $x$- and $y$:

$$F_{1x}+F_{2x}+F_{3x}=0,$$ $$F_{1y}+F_{2y}+F_{3y}=0.$$

Then, we get:

$$30\ N+0\ N+F_{3x}=0,$$ $$F_{3x}=-30\ N,$$ $$0\ N+46\ N+F_{3y}=0,$$ $$F_{3y}=-46\ N.$$

We can find the magnitude of the third force from the Pythagorean theorem:

$$F_3=\sqrt{F_{3x}^2+F_{3y}^2}=\sqrt{(-30\ N)^2+(-46\ N)^2}=55\ N.$$

Let's find the direction of the third force:

$$cos\theta=\dfrac{F_{3x}}{F_3},$$ $$\theta=cos^{-1}(\dfrac{F_{3x}}{F_3}),$$ $$\theta=cos^{-1}(\dfrac{-30\ N}{55\ N})=123^{\circ}.$$

In order to find correct angle we must subtruct the obtained angle from $360^{\circ}$:

$$\theta=360^{\circ}-123^{\circ}=237^{\circ}.$$

To obtaine zero resultunt force, the third force must have the magnitude 55 N and direction $237^{\circ}$(counted conterclockwise from the $x$-axis).