Force Equilibrium on a Particle

← Back to Problems
Dynamics Beginner Newton's Law

Source: Principles of Physics

Problem Sets:

Dynamics set 1009

Problem

A particle moves with a constant velocity $\vec{v}$. Three forces, $\vec{F}_1$, $\vec{F}_2$, and $\vec{F}_3$, act on it. The first two forces are given by $\vec{F}_1 = F_{1x}\hat{i} + F_{1y}\hat{j} + F_{1z}\hat{k}$ and $\vec{F}_2 = F_{2x}\hat{i} + F_{2y}\hat{j} + F_{2z}\hat{k}$.

What is the third force, $\vec{F}_3$?
$\vec{F}_3 = -(F_{1x} + F_{2x})\hat{i} - (F_{1y} + F_{2y})\hat{j} - (F_{1z} + F_{2z})\hat{k}$

Since the particle moves at a constant velocity, its acceleration is zero, $\vec{a} = 0$. According to Newton's second law, the net force on the particle must be zero.

$$\sum \vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = m\vec{a} = 0$$

Solving for the third force $\vec{F}_3$:

$$\vec{F}_3 = -(\vec{F}_1 + \vec{F}_2)$$

The sum of the first two forces is found by adding their respective components:

$$\vec{F}_1 + \vec{F}_2 = (F_{1x} + F_{2x})\hat{i} + (F_{1y} + F_{2y})\hat{j} + (F_{1z} + F_{2z})\hat{k}$$

Therefore, the third force is the negative of this sum.