The angle is 180°, because force of friction always opposes the relative motion.

Since \(tan \theta =\mu =\sqrt{3}\)
\(\theta =tan ^{-1}(\sqrt{3})\Rightarrow \theta =60^o\)

When a person driving a car suddenly applies the brakes, the lower part of the body slower down with the car while upper part of the body continues to move forward due to inertia of motion. If driver is not wearing seat belt, then he falls forward and his head hit against the steering wheel.

 When a lift descending with a downward acceleration a, the apparent weight of a body of mass m is given by w' = R = m (g - a)
Given, Mass of the person, m = 50 kg
Descending acceleration, a = 9 m/s²
Acceleration due to gravity, g = 10 m/s² 
Apparent weight of  the person,
R = m ( g - a ) = 50 ( 10 - 9 ) = 50 N
\(\therefore \)   Reading of the weighing scale = \(\frac { R }{ g } \) = \(\frac { 50 }{ 10 } \) = 5 kg

Given, mass, m = 72.2 kg
Gravity acceleration, g = 9.8 m/s² 
Scale reading = apparent weight = R = ?
(i)  While descending with constant velocity, a = 0
\(\therefore \)    R =  mg
R = 72.2 x 9.8
\(\Longrightarrow \)  R = 707.56 N
(ii)  While ascending with a = 3.2 m/s² 
R = m ( g + a )
R = 72.2 ( 9.8 + 3.2 ) = 938.6 N

Given, F = 36 dyne at an angle of 6000 .
∴∴   Component of force along x-direction
F_x = F cos 6000 = 36 x 1/2 = 18 dyne
 But  F_x = ma_x ,
a_x = \(\frac{F_x}{m} =\frac {18}{18} = 1\) cm/s²

When S ∝ t, so acceleration = 0, Therefore, no external force is acting on the body.

We know that, F = rate of change of momentum
F = 1 kg-m/s² = 1 N

Fiven F = 1N, m = 1 g  = 10^-3 kg
Now, F = ma ⟹ a = \(\frac{F}{m} = \frac{1}{10^{-3}}\)
= 10³ m/s²

According to the Newton's second law of motion. F = ma, for given acceleration a if m is large. F should be more i.e. greater force will be required to put a larger mass in motion.

In solving inclined plane problems, the x and y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are
(i) mg sin \(\alpha\)acting down the plane and
(ii) mg cos \(\alpha\) acting perpendicular to the plane.

N = mg cos \(\alpha\)
mg sin \(\alpha\) -\(\mu\) N =ma
mg sin \(\alpha\) -\(\mu\) mg cos \(\alpha\) =ma
a = g sin \(\alpha\) -\(\mu\)g cos \(\alpha\)
= g (sin \(\alpha\) - \(\mu\)cos \(\alpha\))
= 9.8 ( \({1\over 2}\) - 0.4 x\({\sqrt{3}\over 2}\))
= 4.9 x 0.3072 = 1.505 m/S²
The box accelerates down the plane at 1.505 m/S²

The tension T in the string will provide the necessary centripetal force \({m v^2\over r}\)
\(i.e., T={m v^2\over r}\)
Here m 500 g = \({1\over2}kg;r=1m\)
\(\therefore T={1\over 2}v^2N\)      ..........(1)
There will not be slack if T \(\ge\) weight of the body
i.e., T\(\ge\)mg
\(or {1\over2}v^2\ge{1\over 2}\times 10\)
\(v^2 \ge 10 or v\ge \sqrt{10}ms^{-1}\)
= 3.162 ms^-1.

\(\because \)  Mass of the helicopter, m₁ = 1000 kg
Mass of the crew and the passengers, m₂ = 300 kg
Acceleration of the helicopter, a = 15 m/ s²
Acceleration  due to gravity, g = 10 m/s²
(i) Let R, be the reaction applied by the floor on the crew and the passengers.

R₁- m₂g = m₂a
or
R₁= m₂g + m₂a = m₂(g+a)
= 300(10 + 15) = 7500N(upward direction)
(ii) Action of the rotor of the helicopter on the surrounding air
= ( m₁ + m₂ ) g + ( m₁ + m₂ ) a
= ( m₁ + m₂ ) ( g + a )
= ( 1000 + 300 ) x (10 + 15 )
= 1300 x 25 = 32500 N
Force (action ) of the rotor of the helicopter on the surrounding air = 32500 N (downward)

Given: The mass of the body is 5 kgand the two perpendicular forces are 8 N and 6 N.
The forces on the body are shown below:

The resultant force is given as,
F R = 8 2 + 6 2 = 64+36 =10 N
The tangent of the angle is given as,
tanθ= 6 8 θ= tan ⁻¹ ( 3 4 ) =37°
Newton’s second law of the motion is given as,
F=ma
By substituting the given values in the above expression, we get
10 N=( 5 kg )a a=2 m/ s 2
Thus, the magnitude of acceleration is 2 m/ s 2 and its direction is along the direction of the resultant force. The direction of resultant force is at an angle of 37° with 8 N force.

Given, r =10⁴ km = 10⁴ \(\times \) 1000 m = 10⁷ m,
m = 2500 kg
v = 4 kms^-1 = 4\(\times \)10³ ms^-1
Now,centripeta force is F = \(\frac { m{ v }^{ 2 } }{ r } \) 
F = \(\frac { 2500\times (4\times { 10 }^{ 3 }{ ) }^{ 2 } }{ { 10 }^{ 7 } } =\frac { 2500\times 16\times { 10 }^{ 6 } }{ { 10 }^{ 7 } }\)
\( \\ F=\quad \frac { 250\times 16\times { 10 }^{ 7 } }{ { 10 }^{ 7 } }\)
\(= 4000N\)

Figure shows the various forces acting on the rod AB when it is leaning between the wall and the ground without slipping.
Since the rod is in equilibrium, the net force as well as the net torque on it must each be zero. Considering the forces acting on the rod: we have
\(R_1+(-F')=0 or \ R_1 =\mu_2 R_2\) and \(R_2+F+(-W)=0 or \ R_2 +\mu_1 R_1=W\)

Considering next the moments of all forces about A, we have
R₂ x 0 B = W x 0 N + \(\mu\)₂R₂ x OA  or R₂ x A B cos\(\theta =W \times {AB cos \theta \over 2}+\mu_2 R_2 \times AB sin \theta\) 
\(or \ (R_2 -{W\over 2})cos \theta =(\mu_2 R_2)sin \theta\)
\(or \ tan \theta ={({R_2-{W\over 2}})\over \mu_2 R_2}\)
Now W= R₂ + \(\mu\)₁R₁ = R₂ + \(\mu\)₁ ( \(\mu\)₂ R₂)
=R₂ (1 +\(\mu\)₁\(\mu\)₂)
\(\therefore R_2-W/2 =R_2-{R_2\over 2}(1+\mu_1 \mu_2)\)
\(={R_2 \over 2}-{R_2\over 2}\mu_1 \mu_2={R_2 \over 2}(1-\mu_1 \mu_2)\)
\(\therefore tan \theta ={R_2\over 2}{(1-\mu_1 \mu_2)\over \mu_2 R_2}={(1-\mu_1\mu_2)\over 2\mu_2}\)
or \(\theta= tan^{-1} \{ {(1-\mu_1\mu_2)\over 2\mu_2} \}\)

Considering the mass M₁, since it moves down with uniform speed
T₁= M₁g           ......(i)
For mass M₂, T₁ = T₂+μM₂g cos37⁰ + M₂g sin37⁰    ....(ii)
For mass M₃, T₂ = μM₃g          .....(iii)

Substituting for T₁ and T₂ from (i) and (iii) respectively in (ii),
M₁g = μM₃g + μM₂g cos37⁰ + M₂g sin37⁰
M₁=μM₃ + μM₂ cos37⁰ + M₂ sin37⁰
cos37⁰ =\(\frac{4}{5}\); sin37⁰=\(\frac{3}{5}\); μ=\(\frac{1}{4}\)
∴ M₁ =\(\frac { 1 }{ 4 } \times 4+\frac { 1 }{ 4 } \times 4\times \frac { 4 }{ 5 } +4\times \frac { 3 }{ 5 } \)
=\(1+\frac { 4 }{ 5 } +\frac { 12 }{ 5 } \)=4.2 kg
T₂ = μM₃g =\(\frac{1}{4}\) x 4 x 9.8 = 9.8 N.

(i) a_M = 0.2 m / s², a_W = .25 m
(ii) No change
(iii) a_M = 2,62 m/ s² , a_W = 2 m /s²