Q.Four strings pull on a single point P, which stays at rest (equilibrium). Taking the upward vertical as reference: one string pulls with a force of $2\ \text{N}$ directed up and to the left, at $45^\circ$ from the upward vertical; a second string pulls with $1\ \text{N}$ directed up and to the right, at $45^\circ$ from the upward vertical; a third string pulls with an unknown force $\mathbf{F}_1$ directed horizontally to the right (it makes $45^\circ$ with the $1\ \text{N}$ force); and a fourth string pulls with an unknown force $\mathbf{F}_2$ directed vertically downward (at $90^\circ$ from $\mathbf{F}_1$). Find the forces $\mathbf{F}_1$ and $\mathbf{F}_2$.

Step 1 — Set up the forces at the junction O

As shown in the figure, three forces act at the junction O: the tension $T_1$ in the horizontal string (pulling toward the wall), the tension $T_2$ in the string inclined at $45^\circ$ to the horizontal (pulling toward the ceiling attachment), and the weight $Mg$ of the hanging mass (pulling vertically downward). Resolving $T_2$ into components gives a horizontal part $T_2\cos45^\circ$ and a vertical part $T_2\sin45^\circ$.

Step 2 — Apply equilibrium along each axis

Since O is in equilibrium (the system is static), the net force in both the vertical and horizontal directions must be zero. …

We have a simple pendulum bob with mass $m = 4\ \text{g} = 4 \times 10^{-3}\ \text{kg}$ and charge $q = 20\ \mu\text{C} = 20 \times 10^{-6}\ \text{C}$. It is placed in a uniform horizontal electric field $E = 1000\ \text{V/m}$. At equilibrium, the bob is stationary, so the net force on it must be zero. The forces acting are:

• Weight $mg$ vertically downward.
• Electric force $qE$ horizontally (direction depends on sign of charge; we take it as horizontal).
• Tension $T$ along the string.

At equilibrium, the bob hangs at some angle $\theta$ from the vertical. The tension must balance both the weight and the electric force. This is a classic case of equilibrium under two perpendicular forces: the vertical component of tension balances weight, and the horizontal component balances the electric force.

1. Resolve forces.

   Let $\theta$ be the angle the string makes with the vertical. Then:

   • Vertical: $T \cos\theta = mg$
   • Horizontal: $T \sin\theta = qE$

2. Divide the two equations to eliminate $T$:

3. Plug in the numbers (use SI units):

So

$\tan\theta = \frac{0.02}{0.0392} \approx 0.5102$ …

The key concept here is equilibrium of forces in two dimensions. When the electric field is switched on, the charged ball experiences a horizontal electric force $qE$ and a downward gravitational force $mg$. The string tension adjusts so that the net force is zero. At equilibrium, the string makes a 45° angle with the vertical, which tells us the horizontal and vertical force components are equal in magnitude.

1. Draw the free-body diagram.

   The ball has three forces:

   • Weight $mg$ (downward)
   • Electric force $qE$ (horizontal, direction of the field)
   • Tension $T$ along the string (at 45° to the vertical)

2. Resolve tension into components.

   Since the string makes 45° with the vertical, the horizontal component is $T \sin 45^\circ$ and the vertical component is $T \cos 45^\circ$. Because $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, both components have magnitude $T/\sqrt{2}$.

3. Apply equilibrium conditions.

   • Vertical equilibrium: $T \cos 45^\circ = mg \quad \Rightarrow \quad \frac{T}{\sqrt{2}} = mg$
   • Horizontal equilibrium: $T \sin 45^\circ = qE \quad \Rightarrow \quad \frac{T}{\sqrt{2}} = qE$

4. Equate the two expressions. …

Weight of the stone: $W = mg = 2\times 10 = 20$ N.

Let $T_1$ be the tension in the string at $60^\circ$ and $T_2$ the tension in the string at $30^\circ$ (angles from the horizontal). The $30^\circ$ string is the longer one.

Horizontal balance:

Vertical balance:

$T_1\sin 60^\circ + T_2\sin 30^\circ = 20$ …

Three forces are concurrent at P: $F_A$ along the left string ($30^\circ$ above horizontal, pulling up-left), $F_B$ along the right string ($45^\circ$ above horizontal, pulling up-right), and the weight $F_C=100\ \text{N}$ downward.

Horizontal equilibrium (the two string tensions balance):

$F_B = F_A\,\frac{\cos30^\circ}{\cos45^\circ} = F_A\,\frac{0.866}{0.707} = 1.2247\,F_A$ …

Step 1 — Why a dipole feels a torque but no net force.

In a uniform field, the $+q$ end feels $q\vec{E}$ one way and the $-q$ end feels $q\vec{E}$ the opposite way. The two forces are equal and opposite, so the net force is zero — but because they act along different lines, they form a couple that twists the dipole. The couple's moment is the torque:

where $\theta$ is the angle between the dipole moment $\vec{p}$ and the field $\vec{E}$. Physically, this torque acts to align $\vec{p}$ with $\vec{E}$ — which is why $\tau = 0$ at $\theta = 0^\circ$ (aligned, stable) and is maximum at $\theta = 90^\circ$.

Step 2 — List the data.

Step 3 — Substitute.

First the product $pE$:

Now the trigonometric factor, $\sin 30^\circ = \tfrac{1}{2}$:

Step 4 — Check the distractors. …

Step 1 — Get the direction of the magnetic force.

The force on a current-carrying rod is

Here $\vec{L}$ is horizontal (the rod lies horizontally across the incline) and $\vec{B}$ is vertical. A cross product of a horizontal vector with a vertical vector is horizontal, and perpendicular to the rod.

Since they are mutually perpendicular, the magnitude is simply

acting horizontally, pushing the rod into the incline.

Step 2 — Resolve along the incline (the plane is smooth ⇒ no friction).

Take axes along and perpendicular to the incline surface, $\theta = 30^\circ$:

• Component of weight down the incline: $\;mg\sin\theta$
• Component of the horizontal magnetic force up the incline: $\;BIL\cos\theta$

(The horizontal force makes angle $\theta$ with the incline surface, so its along-incline component carries $\cos\theta$ — this is the step most students get wrong by writing $\sin\theta$.)

Step 3 — Equilibrium condition.

$BIL\cos\theta = mg\sin\theta$ …

The mass is held in equilibrium at $60^\circ$ from the vertical by a horizontal force. Resolving:

Vertical: $T\cos\theta = mg$. …

The concept: apparent weight in an accelerating lift

When you stand in a lift, the floor pushes up on you with a normal force $N$. That’s the force you feel as your weight. If the lift is stationary or moving at constant speed, $N$ equals your true weight $mg$. But if the lift accelerates, the net force on you must equal $ma$ (Newton’s second law). The direction of acceleration decides whether $N$ is larger or smaller than $mg$.

Here the lift is moving down with acceleration $a = 1.8\ \mathrm{m/s^2}$. That means the net acceleration of the man is also $1.8\ \mathrm{m/s^2}$ downward. So the net force on him is $ma$ downward. The two vertical forces acting on him are:

• his weight $mg$ (downward)
• the normal force $N$ from the floor (upward)

Since the net force is downward, the downward force (weight) must be larger than the upward force (normal). So $N < mg$.

Step-by-step solution

1. Choose a sign convention.

Let downward be positive. Then:

• Weight $mg$ is positive.
• Normal force $N$ is negative (it points upward).
• Acceleration $a$ is positive (downward).

2. Write Newton’s second law.

The net force in the downward direction is:

3. Solve for $N$.

Rearrange:

4. Plug in the numbers. …