Q.A rectangular frame is to be suspended symmetrically from overhead supports by two strings of equal length, each string tied to one of the two upper corners of the frame. This can be done in three ways that differ only in how steeply the strings are inclined to the vertical: in arrangement

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. …