Cn u xplain the solutions steps by steps? Cz i really dnt understand wen i revise it nw. Tnxs u kindly. Two particles A & B are connected by a light inextensible string passing over a smooth pulley P. Particle A, of mass m, lies on a plane inclined at 30degree to the horizontal with the part AP of the string parallel to a line of greatest slope. Particle B, of mass 2 m, lies on a horizontal plane with the part BP of the string horizontal & the plane APB vertical. (i) assuming that both planes are smooth, show that when the particles are released from rest the acceleration of B is (1/6)g. Find the expression for the tension in the string in terms of m & g. (ii) the acceleration of B was in fact observed to be (1/8) g.Assuming that the horizontal plane is rough & the inclined plane is smooth, calculate the value of the coefficient of friction between B & the horizontal plane. Answer: (i) (1/3) mg (ii) (1/16)

chngng nms ech tym don't help

Certainly! Let's go through the solution step by step.

(i) First, let's analyze the forces acting on each particle:

For particle A:
- There is its weight, mg, acting vertically downwards.
- There is tension T in the string acting horizontally towards the right.
- There is a normal reaction R perpendicular to the inclined plane.

For particle B:
- There is its weight, 2mg, acting vertically downwards.
- There is tension T in the string acting horizontally towards the left.
- There is a normal reaction R' perpendicular to the horizontal plane.
- There is also a frictional force, F, acting horizontally towards the right.

Since both planes are assumed to be smooth, there is no friction between particle A and the inclined plane.

To determine the acceleration of particle B, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

For particle A:
The net force acting on particle A is given by:
F_net_A = T - mg(sin 30°)

For particle B:
The net force acting on particle B is given by:
F_net_B = 2mg - T

Since the masses are canceled out in the equations, we can solve the system of equations simultaneously to find the value of T.

F_net_A = F_net_B
T - mg(sin 30°) = 2mg - T
2T = 3mg(sin 30°)
T = (3mg/2)(sin 30°)

Now, let's find the acceleration, a, of particle B.
F_net_B = m_B * a
2mg - T = 2m * a
2mg - (3mg/2)(sin 30°) = 2m * a

We know that the acceleration of particle A, a_A, is given by:
a_A = g(sin 30°)

Since the string is inextensible, the accelerations of both particles are the same:
a_A = a_B = a

Therefore,
2mg - (3mg/2)(sin 30°) = 2m * a
2mg - (3mg/2)(1/2) = 2m * a
2mg - (3mg/4) = 2m * a
(8mg - 3mg)/4 = 2m * a
(5mg)/4 = 2m * a
a = (5g)/8

Now, we can substitute the given value of g:
a = (5 * 9.8)/8
a = 49/8 (approx. 6.125) m/s²

Since the acceleration of B is (1/6)g, we can equate the two values of acceleration:
(1/6)g = (5g)/8

Solving for g, we get:
g = (5/8) * (6/1)
g = (5/4) * 3
g = 15/4
g = 3.75 m/s²

Now, let's find the expression for the tension in the string, T, in terms of m and g:

T = (3mg/2)(sin 30°)
T = (3mg/2)(1/2)
T = (3mg)/4

(ii) Now, let's move on to the second part of the question.

In this case, we assume that the horizontal plane is rough, which means there is a frictional force acting on particle B.

The net force acting on particle A remains the same:
F_net_A = T - mg(sin 30°)

For particle B:
The net force acting on particle B is given by:
F_net_B = 2mg - T - F

The frictional force, F, acts in the opposite direction of the motion of particle B.

Since the acceleration of particle B is (1/8)g, we can write:
F_net_B = m_B * a
2mg - T - F = 2m * (1/8)g
2mg - T - F = (m/4)g

Now, let's find the value of the coefficient of friction, μ, between particle B and the horizontal plane:

The frictional force, F, can be expressed as:
F = μR'
where R' is the normal reaction.

The normal reaction can be calculated as:
R' = mg - 2ma

Substituting the expressions for T and R' into the equation:
2mg - (3mg)/4 - μ(mg - 2m(1/8)g) = (m/4)g

Simplifying the equation:
8mg - 3mg - 4μ(mg - m(1/4)g) = mg
8mg - 3mg - 4μ(mg - mg/4) = mg
(32mg - 12mg - 4μmg + μmg) = 4mg
(20mg - 4μmg) = 4mg
20mg = 8μmg
20 = 8μ
μ = 20/8
μ = 2.5

Therefore, the value of the coefficient of friction between particle B and the horizontal plane is 2.5.

I hope this helps! Let me know if you have any further questions.