For Equilibrium Condition,

T1 = 70 N and T2 = 50 N so, T1: T2 = 7 : 5

As the person is in equilibrium net force acting on it should be zero,

so, mg = N + F

⇒ N = mg - F

Weight of block $A$ = $4 \times 10 = 40$ N.

Weight of block $B$ = $2 \times 10 = 20$ N.

Tension in the string $T$ = Weight of $B$ = 20 N.

Normal reaction on $A$, $N =$ Weight of $A$ – Tension $T$ = $40 - 20 = 20$ N.

Maximum friction force acting on the object will be uN= 0.3 * 30 = 9 N

But applied force is 5N so friction force acting on the object will be 5N

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,

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

.

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.

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.