If the angle $\theta$ is with the vertical, then we need to adjust the force components accordingly:

1. Normal Force:

   $$N = mg - F \cos \theta$$

2. Friction Force:

   $$f_s = \mu N = \mu (mg - F \cos \theta)$$

3. Horizontal Component of Force:

   $$F_x = F \sin \theta$$

At the verge of motion:

$$F \sin \theta = \mu (mg - F \cos \theta)$$

Solving for $F$:

$$F = \frac{\mu mg}{\sin \theta + \mu \cos \theta}$$

In translational equilibrium the net force acting on the object is zero. so the object moves with constant velocity.

Given

$m_1 = 5$

kg,

$m_2 = 3$

kg, and

$g = 10$

m/s²:

• Tension:
  $T = m_2 g = 3 \times 10 = 30$ 

  N

• Friction force:
  $f = \mu N = \mu \times 50$ 
• At equilibrium:
  $\mu \times 50 = 30$ 

Solving,

$\mu = \frac{30}{50} = 0.6$

.

\[
\text{Given:} \quad m = 5 \text{ kg}, \quad g = 10 \text{ m/s}^2, \quad \mu = 0.5
\]

\text{Normal force:}
\[
N = F
\]

\text{Friction force:}
\[
f = \mu N = \mu F
\]

\text{For equilibrium:}
\[
\mu F = mg
\]

\text{Substituting values:}
\[
0.5 F = 5 \times 10
\]

\[
0.5 F = 50
\]

\[
F = \frac{50}{0.5} = 100 \text{ N}
\]

\textbf{Final Answer:}
\[
\mathbf{F = 100 N}
\]

The force required to keep the conveyor belt moving at a constant velocity is given by the rate of change of momentum.

Momentum of sand per second:

$$\text{Force} = \frac{\text{dP}}{\text{dt}} = \frac{\text{dm}}{\text{dt}} \times v$$

Given $\frac{dm}{dt} = 2$ kg/s and $v = 4$ m/s,

$$F = 2 \times 4 = 8 \text{ N}$$

Thus, the additional force required is 8 N.

Solution:

Impulse is given by the change in momentum:

$$J = \Delta p = m \Delta v$$

From the graph, we find velocity before and after $t = 2s$:

• Before $t = 2s$:

  $$v_{\text{initial}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{0 - 20}{2 - 0} = -10 \text{ m/s}$$

• After $t = 2s$:

  $$v_{\text{final}} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{20 - 0}{4 - 2} = 10 \text{ m/s}$$

Now, impulse:

$$J=m(v_{\text{final}}-v_{\text{initial}})$$

$$J = m (v_{\text{final}} - v_{\text{initial}})$$ $$J = 2 \times (10 - (-10)) = 2 \times 20 = 40 \text{ Ns}$$

Final Answer:

$$\mathbf{40 \text{ Ns}}$$

For Equilibrium,

2F = Mg ⇒ F = Mg/2

The condition to prevent the block from falling is that the friction force must be at least equal to the weight of the block:

$$f_s \geq mg$$

Since static friction is given by $f_s = \mu N$ and the normal force is due to pseudo force $N = ma$, we get:

$$\mu ma \geq mg$$

Dividing both sides by $\mu m$:

$$a \geq \frac{g}{\mu}$$

Thus, the required acceleration of the cart is $a \geq \frac{g}{\mu}$.

When the truck accelerates to the right with $a$, a pseudo force $ma$ acts to the left on the bob in the truck's frame. The bob reaches equilibrium where the tension components balance forces:

$$\tan \theta = \frac{\text{Pseudo force}}{\text{Weight}} = \frac{ma}{mg} = \frac{a}{g}$$ $$\theta = \tan^{-1} \left(\frac{a}{g} \right)$$

Thus, the pendulum tilts to the left at an angle $\theta = \tan^{-1}(a/g)$ with the vertical.

The forces acting on the body are: Normal reaction (N$=m\mathrm{g)}$(upward) and Friction force f$\le \mu mg$(opposing applied force)

The resultant force is:

$$F = \sqrt{N^2 + f^2} = \sqrt{mg^2 + (\mu mg)^2} = mg \sqrt{1 + \mu^2}$$

$F \leq mg(1 + \mu^2)^{1/2}$