Answer

Verified

341.1k+ views

**Hint:**The given rod is in an equilibrium condition, which means that the net torque and the net force on the rod is zero. Therefore, equate the net torque about the hinges point equal to equal to find the tension. Then equate the net force to zero to find the hinge force.

**Formula used:**

${{F}_{g}}=mg$

$\tau =Fr$

**Complete answer:**

(a) The string will experience a tension in it because the rod will be pulling it downwards due to the gravitational force. Let the tension in the string be T.

We can see that the rod is in equilibrium. This means that the net torque on the rod is zero. Let us analyse the force on the rod.

The earth will exert a gravitational force on the rod in the downward direction. The magnitude of gravitational force on a body is given as ${{F}_{g}}=mg$, where m is the mass of the body and g is acceleration due to gravity. The gravitational force will be at the centre of mass of the body. Therefore, the gravitational force on the rod is acting on the centre of the rod (i.e. centre of mass).

In this case, m = 4kg and $g=10m{{s}^{-2}}$.

${{F}_{g}}=4\times 10=40N$.

Due to the tension in the string, the rod will experience a force as shown. Resolve the tension into its horizontal and vertical components.

The horizontal component of the tension will be $T\cos {{60}^{\circ }}=T\left( \dfrac{1}{2} \right)=\dfrac{T}{2}$.

And its vertical component will be $T\sin {{60}^{\circ }}=T\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\sqrt{3}T}{2}$.

Let us calculate the torque about the hinge point. The magnitude of torque at a point about a its hinged point is given as $\tau =Fr$, where r is the distance between the hinged point and the point of force and F is the force perpendicular to r.

Let us consider the torque in clockwise direction as positive torque and the torque in anticlockwise direction as negative torque.

Therefore, the torque created by the gravitational force is ${{\tau }_{1}}={{F}_{g}}\dfrac{l}{2}=40\times \dfrac{2}{2}=40Nm$.

The torque created by the tension is ${{\tau }_{2}}=-\dfrac{\sqrt{3}T}{2}l=-\dfrac{\sqrt{3}T}{2}(2)=-\sqrt{3}T$.

Hence, the net torque is $\tau ={{\tau }_{1}}-{{\tau }_{2}}=40-\sqrt{3}T$.

But we know that $\tau =0$.

$\Rightarrow 40-\sqrt{3}T=0$

$\Rightarrow T=\dfrac{40}{\sqrt{3}}N$.

(b) The rod is in translational equilibrium too. Hence, the net force on the rod in the horizontal and vertical directions will be zero.

For this, the hinge will apply a force (F) on the rod. Let the horizontal component of the hinge be ${{F}_{x}}$ and its vertical component be ${{F}_{y}}$.

Therefore, the net force in the vertical direction is ${{F}_{y,net}}=-{{F}_{g}}+\dfrac{\sqrt{3}T}{2}+{{F}_{y}}$.

$\Rightarrow {{F}_{y,net}}=-40+\dfrac{\sqrt{3}}{2}\left( \dfrac{40}{\sqrt{3}} \right)+{{F}_{y}}$

But ${{F}_{y,net}}=0$

$\Rightarrow -40+20+{{F}_{y}}=0$

$\Rightarrow {{F}_{y}}=20N$

The net force in the horizontal direction is ${{F}_{x,net}}=-\dfrac{T}{2}+{{F}_{x}}$.

$\Rightarrow {{F}_{x,net}}=-\dfrac{1}{2}\left( \dfrac{40}{\sqrt{3}} \right)+{{F}_{x}}$

But ${{F}_{x,net}}=0$

a

$\Rightarrow {{F}_{x}}=\dfrac{20}{\sqrt{3}}N$

Hence, the force exerted by the hinge is $F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{\dfrac{400}{3}+400}=20\sqrt{\dfrac{4}{3}}=\dfrac{40}{\sqrt{3}}N$.

**Note:**

Remember that we should always check the rotational equilibrium first and then translation equilibrium.In the given case, students may make a mistake by neglecting the rotational equilibrium. If we do that, we will be neglecting the hinge force and the value of the tension in the string will be incorrect.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The branch of science which deals with nature and natural class 10 physics CBSE

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How will you bring about the following conversions class 12 chemistry CBSE

Consider a system of two identical particles One of class 11 physics CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

State the laws of reflection of light