Q: A body of mass \$5\$ kg is acted upon by two perpendicular forces \$8\$ N and \$6\$ N. Given the magnitude and direction of the acceleration of the body.

\$\textbf{Step 1: Drawing FBD}\$ \$\textbf{[Refer Fig. 1]}\$\$\textbf{Step 2: Finding Resultant Force}\$ \$\textbf{[Refer Fig. 2]}\$\$\large \text{Magnitude of Resultant:}\$Since \$F_1\$ and \$F_2\$ are perpendicular\$\therefore \$ Magnitude of Resultant force on body is given by \$R = \sqrt{F_1^2 + F_2^2}\$ \$=\sqrt{8^2+6^2} N\$ \$=10N\$\$\large \text{Direction of Resultant:}\$ \$\theta\ = \tan^{-1} \dfrac{F_2}{F_1}\$ \$\textbf{[Refer Fig. 2]}\$ \$=\tan^{-1} \dfrac{6}{8}\$ \$=36.9^o\$ \$\approx 37^o\$ \$\textbf{Step 3: Applying Newton's second Law for finding acceleration}\$ \$\textbf{[Refer Fig. 3]}\$ \$F_{net} = ma\$ \$\Rightarrow\$ \$10 = 5a\$ \$\Rightarrow\$ \$|\vec{a}| = 2m/s^2\$Hence magnitudes of acceleration is \$2m/s^2\$ directed at \$37^o\$ from the 8 N force.

Q: The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.(i) Angular acceleration(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?(iii) State law of conservation of linear momentum?

\${\omega _1} = 1200\,rpm\$ \$ = {{1200 \times 2\pi } \over {60}} = 40\pi \,rad.{s^{ - 1}}\$ \${\omega _2} = {{3120 \times 2\pi } \over {60}} = 10\pi \,rad.{s^{ - 1}}\$ \$t = 16\,\sec \$ \$(1)\$ Angular acceleration \$a = {{({\omega _2} - {\omega _1})} \over {16}} = {{64\pi } \over {16}}\$ \$ = 4\pi \,rad.{s^{ - 1}}\$ \$(2)\$ Angle, \$\theta\$ \$ = 40\pi \,16 + {1 \over 2}(4\pi ) \times 256\$ \$ = 1152\,rad.\$ \$(3)\$ When there is no any external force applied on a system of partials then, momentum of system of particles will be consecrate or you can say that if external force is zero, then momentum of system of particles is initially \$ = \$ Momentum of system of particles is finally.

Q: A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Mass of car \$m=1800kg\$Distance between front and back axles \$d=1.8m\$ Distance between center of gravity and front axle\$=1.05m\$Let \$R_b\$ and \$R_f\$ be the forces exerted by the level ground on the back and front wheels respectively.At translational equilibrium:\$R_f+R_b=mg=17640\ N.....(i)\$For rotational equilibrium, on taking the torque about the C.G., we have:\$R_f(1.05)=R_b(1.8-1.05)\$\$\implies R_b=1.4R_f.......(ii)\$Solving \$(i)\$ and \$(ii)\$ gives\$R_f=7350\ N\$\$R_b=10290\ N\$The force exerted on each front wheel =\$ 7350 / 2 = 3675\ N\$The force exerted on each back wheel =\$ 10290 / 2 = 5145 N \$

Q: A motorcar is moving with a velocity of \$108\ km/h\$ and it takes \$4\ s\$ to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is \$1000\ kg.\$

Force=mass*acceleration\$f=m \times \dfrac{v-u}{t}\$\$f=1000\times \dfrac{0-108 \times \frac{5}{18}}{4}\$\$=1000 \times (-7.5)\$\$\therefore f=-7500N\$opposing force is denoted by negative sign.

Q: Two equals force are acting at a point with an angle of \$60^{o}\$ between them. If the resultant force is equal to \$40\sqrt {3}\ N\$, the magnitude of each force is

Given,\$F_1=F_2=F(say)\$\$R=40\sqrt{3}N\$\$\theta=60^0\$The Resultant force on the body, \$R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}\$\$40\sqrt{3}=\sqrt{F^2+F^2+2F^2cos60^0}\$\$40\sqrt{3}=\sqrt{2F^2+2F^2\times 0.5}\$\$40\sqrt{3}=\sqrt{3F^2}\$\$40\sqrt{3}=\sqrt{3}F\$\$F=40N\$So, \$F_1=F_2=40N\$The correct option is A.

Q: A piece of wire is bent in the shape of a parabola \$y = kx^{2}\$ (\$y-\$axis vertical) with a bead of mass \$m\$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the \$x-\$axis with a constant acceleration \$a\$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the \$y-\$axis is

\$\textbf{Step 1: Choosing the Frame Of Reference and Drawing the FBD}\$Solving problem from the frame of wire which is accelerating towards right.So, Pseudo force on mass \$m\$ will act on left direction.Pseudo force \$=\$ mass of bead \$\times\$ acceleration of Observer \$=ma\$\$\textbf{Step 2: Equilibrium conditon}\$At equilibrium position, the acceleration of block will be zero in the chosen frame.Therefore, Balancing the forces in x and y direction, we get:\$Nsin\theta=ma\$ \$....(1)\$\$Ncos\theta=mg\$ \$....(2)\$Dividing these two equations, we get\$tan \theta =\dfrac{a}{g}\$ \$....(3)\$\$\textbf{Step 2: Slope of cuve}\$ Equation of curve, \$y = kx^2\$From figure, we see that tangent to the curve at equilibrium point also makes the angle \$\theta with the x axis. Therefore, Slope of curve \$=tan\theta\$\$\Rightarrow\ \ \cfrac{dy}{dx}=tan\theta\$\$\Rightarrow\ \ 2kx=tan\theta\$\$\Rightarrow \ \ 2kx = \dfrac{a}{g}\$ (Using Equation \$3\$)\$\therefore\ \ \displaystyle \mathrm{x}=\frac{\mathrm{a}}{2\mathrm{g}\mathrm{k}} Hence, Option B is correct.

Q: A mass of \$10 kg\$ is suspended by a rope of length \$4 m\$, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of \$45^{\circ}\$ with the vertical. Then F equals: (Take \$g = 10 ms^{-2}\$ and the rope to be massless)

Ref. image I\$m = 10 kg\$\$l = 4 meter\$Ref. image II In \$\Delta ABC\$, \$\angle A + \angle B + \angle C = 180^{\circ}\$\$\Rightarrow 45 + 90 + \angle C = 180^{\circ}\$\$\Rightarrow \angle C = 45^{\circ}\$Ref. image IIISystem is in equilibrium\$\Rightarrow \sum F_y = ma_y = 0\$\$\Rightarrow T_1 \sin 45^{\circ} - mg = 0\$\$\Rightarrow \dfrac{T_1}{\sqrt{2}} = mg \Rightarrow T_1 = mg \sqrt{2}\$ __(I)\$\sum F_x = ma_x = 0\$\$\Rightarrow F - T_1 \cos 45^{\circ} = 0\$\$\Rightarrow F = \dfrac{T_1}{\sqrt{2}} = mg\$ (using eq. (I))\$= 10 \times 10 = 100 N\$

Q: A \$3\ m\$ long ladder weighing \$20\ kg\$ leans on a frictionless wall. Its feet rest on the floor \$1\ m\$ from the wall as shown in Fig. Find the reaction forces of the wall and the floor.

The ladder AB is \$3\$m long, its foot A is at distance AC\$=1\$m from the wall. From Pythagons theorem, \$BC=2\sqrt 2\$m. The forces on the ladder are its weight \$W\$ acting at its centre of gravity \$D\$, reaction forces \$F_1\$ and \$F_2\$ of the wall and the floor respectively. Force \$F_1\$ is perpendicular to the wall, since the wall is frictionless. Force \$F_2\$ is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.For translational equilibrium, taking the forces in the vertical direction,\$N-W=0\$ \$(i)\$Taking the forces in the horizontal direction,\$F-F_1=0\$ \$(ii)\$For rotational equilibrium, taking the moments of the forces about A, \$2\sqrt 2F_1-(1/2)W=0\$ \$(iii)\$Now \$W=20g=20\times 9.8N=196.0N\$From \$(i) N=196.0\$From \$(iii) F_1=W/4\sqrt 2=196.0/4\sqrt 2=34.6N\$From \$(ii) F=F_1=34.6N\$\$F_2=\sqrt{F^2+N^2}=199.0N\$The force \$F_2\$ makes an angle \$\alpha\$ with the horizontal,\$\tan \alpha=N/F=4\sqrt 2, \alpha =\tan^{-1}(4\sqrt 2)=80^o\$

Q: Two forces \${ F }_{ 1 }\$ and \${ F }_{ 2 }\$ are acting on a body. One force is double that of the other force and the resultant is equal the greater force. Then the angle between the two forces is-

Given \$F_1=F\ \ \ \ \ \ F_2=2F\$ and\$F_R=2F\$As we know \$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2cosA}\$putting the above values, we get\$-F^2=4FcosA\$\$cosA=\dfrac{-1}{4}\$\$A=cos^{-1}(\dfrac{-1}{4})\$

Question 1

A body of mass $5$ kg is acted upon by two perpendicular forces $8$ N and $6$ N. Given the magnitude and direction of the acceleration of the body.

Accepted Answer

$\textbf{Step 1: Drawing FBD}$      $\textbf{[Refer Fig. 1]}$$\textbf{Step 2: Finding Resultant Force}$      $\textbf{[Refer Fig. 2]}$$\large \text{Magnitude of Resultant:}$Since $F_1$ and $F_2$ are perpendicular$\therefore $ Magnitude of Resultant force on body is given by                  $R = \sqrt{F_1^2 + F_2^2}$                     $=\sqrt{8^2+6^2} N$ $=10N$$\large \text{Direction of Resultant:}$                $\theta\ = \tan^{-1} \dfrac{F_2}{F_1}$        $\textbf{[Refer Fig. 2]}$                                           $=\tan^{-1} \dfrac{6}{8}$ $=36.9^o$ $\approx 37^o$ $\textbf{Step 3: Applying Newton's second Law for finding acceleration}$      $\textbf{[Refer Fig. 3]}$               $F_{net} = ma$           $\Rightarrow$      $10 = 5a$           $\Rightarrow$       $|\vec{a}| = 2m/s^2$Hence magnitudes of acceleration is $2m/s^2$ directed at $37^o$ from the 8 N force.

Question 2

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.
(i) Angular acceleration
(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?
(iii) State law of conservation of linear momentum?

Accepted Answer

${\omega _1} = 1200\,rpm$&#10;&#10;$ = {{1200 \times 2\pi } \over&#10;{60}} = 40\pi \,rad.{s^{ - 1}}$&#10;&#10;${\omega _2} = {{3120 \times&#10;2\pi } \over {60}} = 10\pi \,rad.{s^{ - 1}}$&#10;&#10;$t = 16\,\sec $&#10;&#10;$(1)$ Angular acceleration&#10;&#10;$a = {{({\omega _2} - {\omega&#10;_1})} \over {16}} = {{64\pi } \over {16}}$&#10;&#10;$ = 4\pi \,rad.{s^{ - 1}}$&#10;&#10;$(2)$ Angle, $\theta$ $ = 40\pi&#10;\,16 + {1 \over 2}(4\pi ) \times 256$&#10;&#10;$ = 1152\,rad.$&#10;&#10;$(3)$ When there is no any&#10;external force applied on a system of partials then, momentum of system of&#10;particles will be consecrate or you can say that if external force is zero,&#10;then momentum of system of particles is initially $ = $ Momentum of system of&#10;particles is finally.

Question 3

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $\vec { F }$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

Accepted Answer

Given: A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $\vec F$  is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. To find the location of P with respect to C.Solution:To have the required motion, the force $\vec F$ has to be applied at the center of mass.The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object.The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is &#8216;m&#8217;, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance &#8216;r&#8217; of the centre of mass of the system from the point C is given by$r=\dfrac {m_1r_1+m_2r_2}{m_1+m_2}\\implies r=\dfrac {m\times 2L+2m\times L}{m+2m}\\implies r=\dfrac {4L}{3}$is the location of P with respect to C.

Question 4

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Accepted Answer

Mass of car $m=1800kg$Distance between front and back axles $d=1.8m$ Distance between center of gravity and front axle$=1.05m$Let $R_b$ and $R_f$ be the forces exerted by the level ground on the back and front wheels respectively.At translational equilibrium:$R_f+R_b=mg=17640\ N.....(i)$For rotational equilibrium, on taking the torque about the C.G., we have:$R_f(1.05)=R_b(1.8-1.05)$$\implies R_b=1.4R_f.......(ii)$Solving $(i)$ and $(ii)$ gives$R_f=7350\ N$$R_b=10290\ N$The force exerted on each front wheel =$ 7350 / 2  =  3675\ N$The force exerted on each back wheel =$ 10290 / 2  =  5145 N $

Question 5

A motorcar is moving with a velocity of $108\ km/h$ and it takes $4\ s$ to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is $1000\ kg.$

Accepted Answer

Force=mass*acceleration$f=m \times \dfrac{v-u}{t}$$f=1000\times \dfrac{0-108 \times \frac{5}{18}}{4}$$=1000 \times (-7.5)$$\therefore f=-7500N$opposing force is denoted by negative sign.

Question 6

Two equals force are acting at a point with an angle of $60^{o}$ between them. If the resultant force is equal to $40\sqrt {3}\ N$, the magnitude of each force is

Accepted Answer

Given,$F_1=F_2=F(say)$$R=40\sqrt{3}N$$\theta=60^0$The Resultant force on the body, $R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$$40\sqrt{3}=\sqrt{F^2+F^2+2F^2cos60^0}$$40\sqrt{3}=\sqrt{2F^2+2F^2\times 0.5}$$40\sqrt{3}=\sqrt{3F^2}$$40\sqrt{3}=\sqrt{3}F$$F=40N$So, $F_1=F_2=40N$The correct option is A.

Question 7

A piece of wire is bent in the shape of a parabola $y = kx^{2}$ ($y-$axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x-$axis with a constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the $y-$axis is

Accepted Answer

$\textbf{Step 1: Choosing the Frame Of Reference and Drawing the FBD}$Solving problem from the frame of wire which is accelerating towards right.So, Pseudo force on mass  $m$ will act on left direction.Pseudo force $=$ mass of bead $\times$ acceleration of Observer                        $=ma$$\textbf{Step 2: Equilibrium conditon}$At equilibrium position, the acceleration of block will be zero in the chosen frame.Therefore, Balancing the forces in x and y direction, we get:$Nsin\theta=ma$               $....(1)$$Ncos\theta=mg$               $....(2)$Dividing these two equations, we get$tan \theta =\dfrac{a}{g}$                    $....(3)$$\textbf{Step 2: Slope of cuve}$ Equation of curve, $y = kx^2$From figure, we see that tangent to the curve at equilibrium point also makes the angle $\theta  with the x axis. Therefore, Slope of curve $=tan\theta\Rightarrow\ \ \cfrac{dy}{dx}=tan\theta\Rightarrow\ \ 2kx=tan\theta\Rightarrow \ \ 2kx = \dfrac{a}{g}$        (Using Equation $3$)$\therefore\ \   \displaystyle \mathrm{x}=\frac{\mathrm{a}}{2\mathrm{g}\mathrm{k}} Hence, Option B is correct.

Question 8

A mass of $10 kg$ is suspended by a rope of length $4 m$, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^{\circ}$ with the vertical. Then F equals: (Take $g = 10 ms^{-2}$ and the rope to be massless)

Accepted Answer

Ref. image I$m = 10 kg$$l = 4 meter$Ref. image II In $\Delta ABC$, $\angle A + \angle B + \angle C = 180^{\circ}$$\Rightarrow 45 + 90 + \angle C = 180^{\circ}$$\Rightarrow \angle C = 45^{\circ}$Ref. image IIISystem is in equilibrium$\Rightarrow \sum F_y = ma_y = 0$$\Rightarrow T_1 \sin 45^{\circ} - mg = 0$$\Rightarrow \dfrac{T_1}{\sqrt{2}} = mg \Rightarrow T_1 = mg \sqrt{2}$ __(I)$\sum F_x = ma_x = 0$$\Rightarrow F - T_1 \cos  45^{\circ} = 0$$\Rightarrow F = \dfrac{T_1}{\sqrt{2}} = mg$     (using eq. (I))$= 10 \times 10 = 100 N$

Question 9

A $3\ m$ long ladder weighing $20\ kg$ leans on a frictionless wall. Its feet rest on the floor $1\ m$ from the wall as shown in Fig. Find the reaction forces of the wall and the floor.

Accepted Answer

The ladder AB is $3$m long, its foot A is at distance AC$=1$m from the wall. From Pythagons theorem, $BC=2\sqrt 2$m. The forces on the ladder are its weight $W$ acting at its centre of gravity $D$, reaction forces $F_1$ and $F_2$ of the wall and the floor respectively. Force $F_1$ is perpendicular to the wall, since the wall is frictionless. Force $F_2$ is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.For translational equilibrium, taking the forces in the vertical direction,$N-W=0$                                       $(i)$Taking the forces in the horizontal direction,$F-F_1=0$                                    $(ii)$For rotational equilibrium, taking the moments of the forces about A, $2\sqrt 2F_1-(1/2)W=0$                                          $(iii)$Now $W=20g=20\times 9.8N=196.0N$From $(i) N=196.0$From $(iii) F_1=W/4\sqrt 2=196.0/4\sqrt 2=34.6N$From $(ii) F=F_1=34.6N$$F_2=\sqrt{F^2+N^2}=199.0N$The force $F_2$ makes an angle $\alpha$ with the horizontal,$\tan \alpha=N/F=4\sqrt 2, \alpha =\tan^{-1}(4\sqrt 2)=80^o$

Question 10

Two forces ${ F }_{ 1 }$ and ${ F }_{ 2 }$ are acting on a body. One force is double that of the other force and the resultant is equal the greater force. Then the angle between the two forces is-

Accepted Answer

Given $F_1=F\ \ \ \ \    \ F_2=2F$  and$F_R=2F$As we know $F_R=\sqrt{F_1^2+F_2^2+2F_1F_2cosA}$putting the above values, we get$-F^2=4FcosA$$cosA=\dfrac{-1}{4}$$A=cos^{-1}(\dfrac{-1}{4})$

Revise with Concepts

Equilibrium of Particle

Concurrent Forces

A body of mass $5$ kg is acted upon by two perpendicular forces $8$ N and $6$ N. Given the magnitude and direction of the acceleration of the body.

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.
(i) Angular acceleration
(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?
(iii) State law of conservation of linear momentum?

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $F$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

A motorcar is moving with a velocity of $108 k m / h$ and it takes $4 s$ to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is $1000 k g .$

Two equals force are acting at a point with an angle of $6 0^{o}$ between them. If the resultant force is equal to $403 N$ , the magnitude of each force is

A mass of $10 k g$ is suspended by a rope of length $4 m$ , from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $4 5^{\circ}$ with the vertical. Then F equals: (Take $g = 10 m s^{- 2}$ and the rope to be massless)

A $3 m$ long ladder weighing $20 k g$ leans on a frictionless wall. Its feet rest on the floor $1 m$ from the wall as shown in Fig. Find the reaction forces of the wall and the floor.

Two forces $F_{1}$ and $F_{2}$ are acting on a body. One force is double that of the other force and the resultant is equal the greater force. Then the angle between the two forces is-

Revise with Concepts

Equilibrium of Particle

Concurrent Forces

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Given the magnitude and direction of the acceleration of the body.

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find. (i) Angular acceleration (ii) Assuming acceleration to be uniform how many rotations does the engine make this bike? (iii) State law of conservation of linear momentum?

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Two equals force are acting at a point with an angle of 60o between them. If the resultant force is equal to 403​ N, the magnitude of each force is

A mass of 10kg is suspended by a rope of length 4m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45∘ with the vertical. Then F equals: (Take g=10ms−2 and the rope to be massless)

A 3 m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig. Find the reaction forces of the wall and the floor.

Two forces F1​ and F2​ are acting on a body. One force is double that of the other force and the resultant is equal the greater force. Then the angle between the two forces is-

A body of mass $5$ kg is acted upon by two perpendicular forces $8$ N and $6$ N. Given the magnitude and direction of the acceleration of the body.

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16sec.? find.
(i) Angular acceleration
(ii) Assuming acceleration to be uniform how many rotations does the engine make this bike?
(iii) State law of conservation of linear momentum?

A T-shaped object with dimensions shown in the figure is lying on a smooth floor. A force $F$ is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

A motorcar is moving with a velocity of $108 k m / h$ and it takes $4 s$ to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is $1000 k g .$

Two equals force are acting at a point with an angle of $6 0^{o}$ between them. If the resultant force is equal to $403 N$ , the magnitude of each force is

A mass of $10 k g$ is suspended by a rope of length $4 m$ , from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $4 5^{\circ}$ with the vertical. Then F equals: (Take $g = 10 m s^{- 2}$ and the rope to be massless)

A $3 m$ long ladder weighing $20 k g$ leans on a frictionless wall. Its feet rest on the floor $1 m$ from the wall as shown in Fig. Find the reaction forces of the wall and the floor.

Two forces $F_{1}$ and $F_{2}$ are acting on a body. One force is double that of the other force and the resultant is equal the greater force. Then the angle between the two forces is-