Q.There are three forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.

The body moves with uniform speed, which means its acceleration is zero. According to Newton's Second Law, this implies the net force acting on the body is zero.

(a) Concept: Translational Equilibrium and Vector Addition.

Since the body moves with uniform speed, its acceleration is zero. Therefore, the net force acting on it must be zero:

$\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$ …

When a body moves with uniform speed, it means its velocity is constant. This condition is crucial because it directly tells us about the net force and net torque acting on the body.

Concept and Intuition

1. Newton's First Law (or Second Law in equilibrium): A body moving with uniform velocity (constant speed in a straight line, or constant speed if the problem implies no change in direction) has zero acceleration. According to Newton's Second Law, $\mathbf{F}_{net} = m\mathbf{a}$. If $\mathbf{a} = \mathbf{0}$, then the net force $\mathbf{F}_{net}$ acting on the body must be zero. This is a state of translational equilibrium.
2. Vector Addition and Coplanarity: If the vector sum of three forces is zero, these three vectors, when placed head-to-tail, must form a closed triangle. A triangle, by definition, lies entirely within a single plane. Therefore, the three forces must be coplanar.
3. Torque: Torque is the rotational equivalent of force. It measures the tendency of a force to cause rotation about an axis or a point. It is defined as $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, where $\mathbf{r}$ is the position vector from the point about which torque is calculated to the point of application of the force $\mathbf{F}$. For a body to be in rotational equilibrium (i.e., not accelerating rotationally), the net torque acting on it must be zero.

Step-by-Step Solution

(a) Showing that the forces are coplanar

1. Condition for uniform speed: The problem states that the body moves with uniform speed. This implies that its acceleration is zero ($\mathbf{a} = \mathbf{0}$).

   Important

   For a body moving with uniform speed, its acceleration is zero. By Newton's Second Law, $\mathbf{F}_{net} = m\mathbf{a}$, so the net force acting on the body must be zero.

   $\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$

2. Geometric interpretation of zero vector sum: When the vector sum of three forces is zero, it means that if you place the vectors head-to-tail, they form a closed polygon. For three vectors, this polygon is a triangle.

   • Imagine placing the tail of $\mathbf{F}_2$ at the head of $\mathbf{F}_1$.
   • Then, place the tail of $\mathbf{F}_3$ at the head of $\mathbf{F}_2$.
   • Since $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$, the head of $\mathbf{F}_3$ must coincide with the tail of $\mathbf{F}_1$.
   • This forms a closed triangle.

3. Conclusion on coplanarity: A triangle, by its very nature, lies entirely within a single plane. Therefore, the three force vectors $\mathbf{F}_1$, $\mathbf{F}_2$, and $\mathbf{F}_3$ must lie in the same plane. Hence, they are coplanar.

(b) Showing that the torque acting on the body about any point due to these three forces is zero

1. Definition of torque: The torque $\boldsymbol{\tau}$ produced by a force $\mathbf{F}$ about a point O is given by the cross product of the position vector $\mathbf{r}$ (from O to the point of application of the force) and the force vector $\mathbf{F}$.

   The torque $\boldsymbol{\tau}$ about a point O due to a force $\mathbf{F}$ applied at point P is:

   $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$

   where $\mathbf{r}$ is the position vector from O to P.

2. Forces act at a single point P: All three forces $\mathbf{F}_1$, $\mathbf{F}_2$, and $\mathbf{F}_3$ act at a common point P on the body. This is a crucial piece of information.

3. Net torque about the point of application P: Let's first calculate the net torque about the point P itself.

   • For any force $\mathbf{F}_i$ acting at point P, the position vector $\mathbf{r}$ from P to the point of application (which is P) is the zero vector, $\mathbf{r} = \mathbf{0}$.
   • Therefore, the torque due to each force about point P is $\boldsymbol{\tau}_i = \mathbf{0} \times \mathbf{F}_i = \mathbf{0}$.
   • The net torque about P is $\boldsymbol{\tau}_{net, P} = \boldsymbol{\tau}_1 + \boldsymbol{\tau}_2 + \boldsymbol{\tau}_3 = \mathbf{0} + \mathbf{0} + \mathbf{0} = \mathbf{0}$.

4. Net torque about any arbitrary point O: Now, consider an arbitrary point O in space. Let $\mathbf{r}_P$ be the position vector from O to the point P where the forces are applied. …

Concept: Equilibrium Under Concurrent Forces

Uniform speed means zero acceleration, so the net force is zero; a zero vector sum of three vectors forces them into a closed (planar) triangle, and their common point of application makes the net torque about any point vanish too.

Step 1: Apply Newton's second law

Uniform speed $\Rightarrow \mathbf a=0 \Rightarrow$ net force $\mathbf F_1+\mathbf F_2+\mathbf F_3=\mathbf 0$.

Step 2: Interpret the zero vector sum geometrically

Placing the three vectors head-to-tail, a zero sum means the chain closes back on its starting point — forming a triangle, which by definition lies in a single plane. Hence the forces are coplanar.

Step 3: Compute the torque about the common point P

Since all three forces act at the same point P, the position vector from P to the point of application is $\mathbf r=\mathbf 0$ for each, so each individual torque about P is zero.

Step 4: Extend to an arbitrary point O …