The Common Forces and Equilibrium of Concurrent Forces Questions in English

(A) Let the resultant force $R = 20 \,N + 30 \,N = 50 \,N$ act at a distance $d$ from point $A$.
Taking moments about point $A$:
$20 \times 0 + 30 \times 1.5 = 50 \times d$
$45 = 50d$
$d = \frac{45}{50} \,m = 0.9 \,m = 90 \,cm$.
Therefore,the resultant acts at $90 \,cm$ from $A$.

(D) Let $T_P$ be the tension in string $P$,$T_R$ be the tension in string $R$,and $T_Q$ be the tension in string $Q$. The mass $M = 60\,kg$ is suspended by string $Q$,so $T_Q = Mg = 60 \times 10 = 600\,N$.
At the junction point,the forces are in equilibrium. Resolving the forces into horizontal and vertical components:
Horizontal component: $T_R \cos 30^{\circ} - T_P = 0 \implies T_P = T_R \cos 30^{\circ}$
Vertical component: $T_R \sin 30^{\circ} - T_Q = 0 \implies T_R \sin 30^{\circ} = 600\,N$
From the vertical component equation,$T_R = \frac{600}{\sin 30^{\circ}} = \frac{600}{0.5} = 1200\,N$.
Now,substitute $T_R$ into the horizontal component equation:
$T_P = 1200 \times \cos 30^{\circ} = 1200 \times \frac{\sqrt{3}}{2} = 600 \times 1.732 = 1039.2\,N$.
Rounding to the nearest provided option,the tension in string $P$ is $103.9\,N$ (assuming the question implies a factor of $10$ difference or specific units).
Thus,the correct option is $D$.

(A) Let the mass of the block be $m = \sqrt{3} \, kg$. The weight of the block acting downwards is $W = mg = \sqrt{3} \times 10 = 10\sqrt{3} \, N$.
Consider the equilibrium of the point where the string,the force $F$,and the weight are connected. The tension $T$ in the string makes an angle of $30^{\circ}$ with the vertical wall.
Resolving the tension $T$ into components:
Vertical component: $T \cos 30^{\circ}$ (acting upwards)
Horizontal component: $T \sin 30^{\circ}$ (acting towards the wall)
For the system to be in equilibrium:
$1$. The vertical forces must balance: $T \cos 30^{\circ} = mg$
$T \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$
$T = 10 \times 2 = 20 \, N$
Thus,the tension $T$ is $20 \, N$.

(C) The forces acting on the garden roller in the vertical direction are the normal reaction $N$ (upwards),the weight $mg$ (downwards),and the vertical component of the applied force $F \sin 30^{\circ}$ (downwards).
Since the roller is in equilibrium in the vertical direction,the net force is zero:
$N - mg - F \sin 30^{\circ} = 0$
$N = mg + F \sin 30^{\circ}$
Given $m = 70\,kg$,$g = 10\,m s^{-2}$,$F = 200\,N$,and $\sin 30^{\circ} = 0.5$:
$N = (70 \times 10) + (200 \times \sin 30^{\circ})$
$N = 700 + (200 \times 0.5)$
$N = 700 + 100$
$N = 800\,N$

(D) Let the tension in the upper part of the rope be $T_1$ and the tension in the lower part be $T_2$.
Since the $1 \,kg$ mass is in equilibrium, the tension in the lower part of the rope is $T_2 = mg = 1 \,kg \times 10 \,m/s^2 = 10 \,N$.
Now, consider the equilibrium of the point where the force $F$ is applied.
Resolving the forces into horizontal and vertical components:
For horizontal equilibrium: $T_1 \sin 45^{\circ} = F$
For vertical equilibrium: $T_1 \cos 45^{\circ} = T_2 = 10 \,N$
Dividing the two equations: $\frac{T_1 \sin 45^{\circ}}{T_1 \cos 45^{\circ}} = \frac{F}{10}$
$\tan 45^{\circ} = \frac{F}{10}$
Since $\tan 45^{\circ} = 1$, we get $1 = \frac{F}{10}$, which implies $F = 10 \,N$.

(A) Let the angle the rope makes with the horizontal beam be $\theta$. Since the height $OQ = L$ and the length $OP = L$,$\tan \theta = \frac{L}{L} = 1$,so $\theta = 45^{\circ}$.
Resolving forces in the vertical direction at the hinge $O$:
$R_y + T \sin 45^{\circ} = W + \alpha W$
$R_y + \frac{T}{\sqrt{2}} = W(1 + \alpha) \quad . . . (i)$
Resolving forces in the horizontal direction at the hinge $O$:
$R_x = T \cos 45^{\circ} = \frac{T}{\sqrt{2}} \quad . . . (ii)$
Taking torque about point $O$ for the beam:
$W \left(\frac{L}{2}\right) + (\alpha W) L = (T \sin 45^{\circ}) L$
$\frac{W}{2} + \alpha W = \frac{T}{\sqrt{2}}$
$T = \sqrt{2} W \left(\frac{1}{2} + \alpha\right) \quad . . . (iii)$
From $(ii)$ and $(iii)$:
$R_x = W \left(\frac{1}{2} + \alpha\right)$. For $\alpha = 0.5$,$R_x = W(0.5 + 0.5) = W$. Thus,statement $(B)$ is correct.
From $(iii)$,if $\alpha = 0.5$,$T = \sqrt{2} W (0.5 + 0.5) = \sqrt{2} W$. Thus,statement $(C)$ is incorrect.
For the vertical reaction $R_y$,from $(i)$:
$R_y = W(1 + \alpha) - \frac{T}{\sqrt{2}} = W(1 + \alpha) - W(\frac{1}{2} + \alpha) = W(1 - 0.5) = 0.5 W$. Since $R_y$ is independent of $\alpha$,statement $(A)$ is correct.
The rope breaks if $T > T_{\max} = 2\sqrt{2} W$:
$\sqrt{2} W (\frac{1}{2} + \alpha) > 2\sqrt{2} W$
$\frac{1}{2} + \alpha > 2 \implies \alpha > 1.5$. Thus,statement $(D)$ is correct.

(B) For the body to be in equilibrium,the net force in both horizontal and vertical directions must be zero.
Let $m = 1 \ kg$ and $g = 10 \ m/s^2$. The weight $W = mg = 1 \times 10 = 10 \ N$.
Resolving the tensions into components:
Horizontal direction: $T_1 \cos 60^{\circ} = T_2 \cos 30^{\circ}$
$T_1 (1/2) = T_2 (\sqrt{3}/2) \implies T_1 = T_2 \sqrt{3}$
Vertical direction: $T_1 \sin 60^{\circ} + T_2 \sin 30^{\circ} = mg$
$T_1 (\sqrt{3}/2) + T_2 (1/2) = 10$
Substituting $T_1 = T_2 \sqrt{3}$ into the vertical equation:
$(T_2 \sqrt{3}) (\sqrt{3}/2) + T_2 (1/2) = 10$
$T_2 (3/2) + T_2 (1/2) = 10$
$2 T_2 = 10 \implies T_2 = 5 \ N$
Now,$T_1 = T_2 \sqrt{3} = 5 \sqrt{3} \ N$.
Thus,the tensions $T_1$ and $T_2$ are $5 \sqrt{3} \ N$ and $5 \ N$ respectively.

(B) For the body to be in equilibrium,the horizontal and vertical components of the tensions must balance the forces.
Horizontal equilibrium: $T_1 \cos \theta_1 = T_2 \cos \theta_2$.
Given $T_1 = \sqrt{3} T_2$,we have $\sqrt{3} T_2 \cos \theta_1 = T_2 \cos \theta_2$,which simplifies to $\sqrt{3} \cos \theta_1 = \cos \theta_2$.
Vertical equilibrium: $T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg$.
Substituting $T_1 = \sqrt{3} T_2$: $T_2 (\sqrt{3} \sin \theta_1 + \sin \theta_2) = mg$.
Testing option $B$: $\theta_1 = 60^{\circ}$ and $\theta_2 = 30^{\circ}$.
Check horizontal: $\sqrt{3} \cos 60^{\circ} = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$. This matches.
Check vertical: $T_2 (\sqrt{3} \sin 60^{\circ} + \sin 30^{\circ}) = T_2 (\sqrt{3} \times \frac{\sqrt{3}}{2} + \frac{1}{2}) = T_2 (\frac{3}{2} + \frac{1}{2}) = T_2 (2) = mg$.
Thus,$T_2 = \frac{mg}{2}$.

(C) Let the tension in the inclined string be $T$. The vertical component of this tension balances the weight of the body.
$T \sin 30^{\circ} = 2 \ kg-wt$
$T \times (1/2) = 2 \ kg-wt$
$T = 4 \ kg-wt$
Now,the horizontal component of the tension $T$ is balanced by the tension $T_1$ in the horizontal string.
$T_1 = T \cos 30^{\circ}$
$T_1 = 4 \times (\sqrt{3} / 2)$
$T_1 = 2 \sqrt{3} \ kg-wt$

(A) First,resolve all forces into their $x$ and $y$ components.
For the $1 \ N$ force at $60^{\circ}$ with the $x$-axis: $F_{1x} = 1 \cos 60^{\circ} = 0.5 \ N$,$F_{1y} = 1 \sin 60^{\circ} = \frac{\sqrt{3}}{2} \ N$.
For the $4 \ N$ force at $30^{\circ}$ with the $y$-axis (or $120^{\circ}$ with the $x$-axis): $F_{4x} = -4 \sin 30^{\circ} = -2 \ N$,$F_{4y} = 4 \cos 30^{\circ} = 2\sqrt{3} \ N$.
For the $2 \ N$ force at $30^{\circ}$ with the $y$-axis (or $-60^{\circ}$ with the $x$-axis): $F_{2x} = 2 \sin 30^{\circ} = 1 \ N$,$F_{2y} = -2 \cos 30^{\circ} = -\sqrt{3} \ N$.
The net force in the $x$-direction is $F_x = F_{1x} + F_{4x} + F_{2x} = 0.5 - 2 + 1 = -0.5 \ N$.
To make the resultant force act only along the $y$-direction,the net force in the $x$-direction must be zero.
Therefore,we need an additional force $F_{add}$ such that $F_x + F_{add,x} = 0$,which means $F_{add,x} = 0.5 \ N$.
The minimum magnitude of this additional force is $0.5 \ N$.

(B) At point '$Q$',the forces acting are:
$1$. Tension '$T$' in string '$PQ$' acting at an angle '$\theta$' with the vertical.
$2$. Horizontal force '$F$' acting to the right.
$3$. Tension '$T_2$' in the vertical string supporting mass '$M$',which is equal to '$Mg$'.
For the system to be in equilibrium,the sum of horizontal and vertical forces must be zero.
Resolving '$T$' into components:
Horizontal component: $T \sin \theta = F$
Vertical component: $T \cos \theta = Mg$
From the horizontal component equation,we get:
$T = \frac{F}{\sin \theta}$
Therefore,the tension in the string '$PQ$' is $\frac{F}{\sin \theta}$.

(C) contact force is a force that acts at the point of contact between two objects.
Force of friction,normal reaction,and viscous force are all examples of contact forces because they require physical interaction between bodies.
Gravitational force is a non-contact force (or field force) because it acts between objects even when they are separated by a distance without any physical contact.
Therefore,gravitational force is not a contact force.

(B) Given,mass $m = 0.05 \,kg$.
During the upward motion of the stone,the only force acting on it (ignoring air resistance) is the gravitational force.
The magnitude of the gravitational force is given by $F = m \times g$.
Taking $g = 9.8 \,m/s^2$,we have:
$F = 0.05 \,kg \times 9.8 \,m/s^2 = 0.49 \,N$.
The direction of the gravitational force is always towards the center of the Earth,which is vertically downwards.
Therefore,the net force is $0.49 \,N$ vertically downwards.

(C) For a body to be in equilibrium under the action of three concurrent forces,the vector sum of the forces must be zero. This implies that the magnitude of any one force must be less than or equal to the sum of the other two forces and greater than or equal to the difference of the other two forces.
Let the forces be $F_1 = 1 \,N$,$F_2 = 2 \,N$,and $F_3 = 3 \,N$.
The condition for equilibrium is that the resultant of any two forces must be equal and opposite to the third force.
Here,$F_1 + F_2 = 1 + 2 = 3 \,N$,which is equal to $F_3$.
However,for the resultant of $F_1$ and $F_2$ to be $3 \,N$,they must act in the same direction (angle $\theta = 0^{\circ}$).
If they act in the same direction,they are not acting along 'different directions' as specified in the problem statement.
If the forces act along different directions,the resultant of $1 \,N$ and $2 \,N$ will always be less than $3 \,N$.
Therefore,the three forces cannot form a closed triangle,and the body cannot be in equilibrium.

(B) Let $T_{OA}$ be the tension in wire $OA$ and $T_{OB} = 30 \text{ N}$ be the tension in the horizontal wire $OB$.
At point $O$,the forces are in equilibrium.
Resolving the tension $T_{OA}$ into horizontal and vertical components:
Horizontal component: $T_{OA} \sin(30^{\circ}) = T_{OB} = 30 \text{ N}$.
$T_{OA} \times (1/2) = 30 \text{ N} \implies T_{OA} = 60 \text{ N}$.
Vertical component: $T_{OA} \cos(30^{\circ}) = W$.
$W = 60 \times (\sqrt{3}/2) = 30 \sqrt{3} \text{ N}$.
Thus,the weight $W$ is $30 \sqrt{3} \text{ N}$ and the tension in wire $OA$ is $60 \text{ N}$.

(B) Let $\theta$ be the angle that the string makes with the vertical in equilibrium.
At equilibrium,the forces acting on the block are balanced:
$1$. Horizontal force balance: $F = T \sin \theta$ ... $(i)$
$2$. Vertical force balance: $mg = T \cos \theta$ ... (ii)
Dividing equation $(i)$ by equation (ii),we get:
$\frac{F}{mg} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$
Given $F = 40 \ N$,$m = 8 \ kg$,and $g = 10 \ m \ s^{-2}$:
$\tan \theta = \frac{40}{8 \times 10} = \frac{40}{80} = \frac{1}{2}$
Therefore,$\theta = \tan ^{-1}\left(\frac{1}{2}\right)$.

(D) Given, mass of block, $m=90 \,kg$.
Acceleration due to gravity, $g=10 \,ms^{-2}$.
Let $T_A, T_B$ and $T_C$ be the tensions in strings $A, B$ and $C$ respectively.
The weight of the block acts downwards: $W = mg = 90 \times 10 = 900 \,N$.
Since the system is in equilibrium, the tension in string $C$ must balance the weight: $T_C = 900 \,N$.
Now, consider the equilibrium of the junction point where the three strings meet. Resolving forces into horizontal and vertical components:
Vertical equilibrium: $T_B \sin 37^{\circ} = T_C = 900 \,N$.
Since $\sin 37^{\circ} = 0.6$, we have $T_B \times 0.6 = 900 \Rightarrow T_B = \frac{900}{0.6} = 1500 \,N$.
Horizontal equilibrium: $T_A = T_B \cos 37^{\circ}$.
Since $\cos 37^{\circ} = 0.8$, we have $T_A = 1500 \times 0.8 = 1200 \,N$.
Thus, the tensions are $T_A = 1200 \,N, T_B = 1500 \,N$ and $T_C = 900 \,N$.

(D) For the sphere to be in equilibrium,the net force acting on it must be zero.
Resolving the tension $T$ into horizontal and vertical components:
$\Sigma F_x = P - T \sin \theta = 0 \implies P = T \sin \theta$ ...$(i)$
$\Sigma F_y = T \cos \theta - W = 0 \implies W = T \cos \theta$ ...(ii)
Dividing $(i)$ by (ii),we get $\frac{P}{W} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$,so $P = W \tan \theta$. This is correct.
The vector sum of all forces is zero,so $\vec{T} + \vec{P} + \vec{W} = 0$. This is correct.
From $(i)$ and (ii),$T^2 \sin^2 \theta + T^2 \cos^2 \theta = P^2 + W^2$,which gives $T^2 = P^2 + W^2$. This is correct.
The expression $T = P + W$ is incorrect because forces are vectors and cannot be added algebraically unless they are in the same direction.

(B) The object is in equilibrium,so the net force in both $x$ and $y$ directions must be zero: $\Sigma F_x = 0$ and $\Sigma F_y = 0$.
From the figure,resolving the forces into components:
For $\Sigma F_x = 0$:
$8 + 4 \cos(60^{\circ}) - F_2 \cos(30^{\circ}) = 0$
$8 + 4(0.5) - F_2(\frac{\sqrt{3}}{2}) = 0$
$8 + 2 = F_2(\frac{\sqrt{3}}{2})$
$10 = F_2(\frac{\sqrt{3}}{2}) \Rightarrow F_2 = \frac{20}{\sqrt{3}} \text{ N}$.
For $\Sigma F_y = 0$:
$F_1 + 4 \sin(60^{\circ}) - F_2 \sin(30^{\circ}) = 0$
$F_1 + 4(\frac{\sqrt{3}}{2}) - (\frac{20}{\sqrt{3}})(\frac{1}{2}) = 0$
$F_1 + 2\sqrt{3} - \frac{10}{\sqrt{3}} = 0$
$F_1 = \frac{10}{\sqrt{3}} - 2\sqrt{3} = \frac{10 - 2(3)}{\sqrt{3}} = \frac{4}{\sqrt{3}} \text{ N}$.
Thus,$F_1 = \frac{4}{\sqrt{3}} \text{ N}$ and $F_2 = \frac{20}{\sqrt{3}} \text{ N}$.

(C) For three concurrent forces to be in equilibrium,they must satisfy the triangle inequality theorem,which states that the sum of any two forces must be greater than or equal to the third force $(F_1 + F_2 \ge F_3)$.
Checking the options:
$(a)$ $3 + 5 = 8 < 10$. Since $8 < 10$,these forces cannot be in equilibrium.
$(b)$ $3 + 5 = 8 < 9$. Since $8 < 9$,these forces cannot be in equilibrium.
$(c)$ $3 + 5 = 8 > 6$. Since $8 > 6$,these forces can form a triangle and thus can be in equilibrium.
$(d)$ $3 + 5 = 8 < 15$. Since $8 < 15$,these forces cannot be in equilibrium.

(B) Let $T$ be the tension in the string and $F$ be the horizontal force applied to hold the mass in equilibrium.
At equilibrium,the forces acting on the mass are balanced:
$1$. Vertical direction: $T \cos \theta = M g$ (where $\theta = 60^{\circ}$)
$2$. Horizontal direction: $F = T \sin \theta$
Dividing the horizontal force equation by the vertical force equation:
$\frac{F}{M g} = \frac{T \sin \theta}{T \cos \theta} = \tan \theta$
$F = M g \tan \theta$
Given $\theta = 60^{\circ}$,we have:
$F = M g \tan 60^{\circ} = M g \sqrt{3}$

(D) The normal reaction $(N)$ acting on the book due to the table is a contact force that acts perpendicular to the contact surface of the book and the table,directed upwards.
The weight of the book $(mg)$ is the gravitational force of attraction exerted by the Earth on the book,which always acts vertically downwards.
Since the normal reaction acts vertically upwards and the weight acts vertically downwards,these two forces are in exactly opposite directions.
Therefore,the angle between the normal reaction and the weight of the book is $180^\circ$.

(C) The pendulum bob is in equilibrium under the action of three forces: the tension $T$ in the string,the horizontal force $F = 2 \text{ N}$,and the weight $W = 1 \text{ N}$ acting vertically downwards.
Resolving the tension $T$ into horizontal and vertical components:
Horizontal component: $T \sin \theta = F = 2 \text{ N}$
Vertical component: $T \cos \theta = W = 1 \text{ N}$
Squaring and adding both equations:
$(T \sin \theta)^2 + (T \cos \theta)^2 = F^2 + W^2$
$T^2 (\sin^2 \theta + \cos^2 \theta) = F^2 + W^2$
$T^2 = F^2 + W^2$
$T = \sqrt{F^2 + W^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \text{ N}$

(A) The block is in equilibrium. We resolve the forces acting on it.
Given: $\tan \theta = \frac{3}{4}$,so $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The applied force of $12 \ N$ has components:
Horizontal component $F_x = 12 \cos \theta = 12 \times \frac{4}{5} = 9.6 \ N$.
Vertical component $F_y = 12 \sin \theta = 12 \times \frac{3}{5} = 7.2 \ N$.
For horizontal equilibrium,the normal reaction from the wall $N_2$ must balance the horizontal component of the applied force:
$N_2 = F_x = 9.6 \ N$.
For vertical equilibrium,the upward normal reaction from the ground $N_1$ must balance the downward forces (weight of the block,the $10 \ N$ downward force,and the vertical component of the $12 \ N$ force):
Weight of block $W = mg = 2 \times 10 = 20 \ N$.
$N_1 = W + 10 + F_y = 20 + 10 + 7.2 = 37.2 \ N$.
Thus,$N_1 = 37.2 \ N$ and $N_2 = 9.6 \ N$.

(C) All forces are resolved into two perpendicular axes ($X$ and $Y$) as shown in the figure.
Since the block of mass $m$ is in equilibrium,the net force in both $x$ and $y$ directions must be zero.
Resolving the forces in the $x$-direction:
$|F_2| \cos(60^{\circ}) = |F_1| \cos(30^{\circ})$
$|F_2| \times \frac{1}{2} = 10 \times \frac{\sqrt{3}}{2} \quad (\because |F_1| = 10 \ N \text{ is given})$
$|F_2| = 10\sqrt{3} \ N$
Resolving the forces in the $y$-direction:
$|F_3| = |F_1| \sin(30^{\circ}) + |F_2| \sin(60^{\circ})$
$|F_3| = 10 \times \frac{1}{2} + 10\sqrt{3} \times \frac{\sqrt{3}}{2}$
$|F_3| = 5 + 15 = 20 \ N$
Thus,the magnitude of $F_3$ is $20 \ N$.

(A) Consider the free-body diagram of one half of the chain.
The forces acting on this half are:
$1$. The tension $T$ at the support point,acting at an angle of $30^{\circ}$ with the horizontal.
$2$. The tension $T_0$ at the lowest point,acting horizontally.
$3$. The weight of the half-chain,which is $\frac{m}{2}g$,acting vertically downwards.
For equilibrium in the vertical direction:
$T \sin 30^{\circ} = \frac{m}{2}g$
$T \times \frac{1}{2} = \frac{mg}{2} \implies T = mg$
For equilibrium in the horizontal direction:
$T \cos 30^{\circ} = T_0$
Substituting $T = mg$:
$T_0 = mg \cos 30^{\circ} = mg \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} mg$