Name: Collision Between a Point Mass and a Rigid Body
Uploaded: 2019-07-19
Duration: 7 min 57 s
Description: Can we treat the collision of a point mass with a rigid body in the same way we treated the collision of point masses? Watch this video to know about it.

Question

In the system as shown in figure, $AB$ is a uniform rod of mass $10 kg$ and $BC$ is a light string which is connected between wall and rod, in vertical plane. There is block of mass $15 kg$ connected at $B$ with a light string. $[Take g = 10 m s^{- 2}]$ ( $BC$ and $BD$ are two different strings)

$(i)$ Tension in the string $BC$

$(ii)$ Hinge force exerted on beam at point $A$

Accepted Answer

In the given case, the entire system is in translational and rotational equilibrium. From the free-body diagram, we know that

Length of rod, $l = 1 m$

Mass of rod, $m_{1} = 10 kg$

Mass of block, $m_{2} = 15 kg$

Let's assume the hinge force which acts in the form of normal force be $N$ . There horizontal and vertical components be $N_{x}$ and $BD$ . And let the tension in string $B C$ be $T$ .

So, for horizontal translational equilibrium to exist, the net force in horizontal direction needs to be zero. That is

$N_{x} - T \cos (53) = 0$

$\Rightarrow N_{x} = T \cos (53) . . . (1)$

Similarly, for vertical translational equilibrium to exist, the net force in vertical direction needs to be zero. That is

$N_{y} + T \sin (53) - m_{1} g - m_{2} g = 0$

$\Rightarrow N_{y} + T \sin (53) = (m_{1} + m_{2}) g . . . (2)$

Now for rotational equilibrium to exist, the net torque about any point in the system must be zero. So, net torque about point $A$ ,

$m_{1} g \times \frac{l}{2} (- \hat{k}) + m_{2} g \times l (- \hat{k}) + T \sin (53) \times l (\hat{k}) = 0$

$\Rightarrow T l \sin (53) - \frac{m_{1} g l}{2} - m_{2} g l = 0$

$\Rightarrow T \times 1 \times \frac{4}{5} = 10 \times 10 \times \frac{1}{2} + 15 \times 10 \times 1$

$\Rightarrow \frac{4 T}{5} = 200$

$\Rightarrow T = 250 N$

Now by using equation $(1)$ ,

$N_{x} = T \cos (53)$

$\Rightarrow N_{x} = 250 \times \frac{3}{5}$

$\Rightarrow N_{x} = 150 N$

And by using equation $(2)$ ,

$N_{y} + T \sin (53) = (m_{1} + m_{2}) g$

$\Rightarrow N_{y} + 250 \times \frac{4}{5} = (10 + 15) 10$

$\Rightarrow N_{y} = 50 N$

So, the resultant normal force, $N = \sqrt{{(N_{x})}^{2} + {(N_{y})}^{2}}$

$\Rightarrow N = \sqrt{{(150)}^{2} + {(50)}^{2}}$

$\Rightarrow N = 50 \sqrt{10} N = 158.11 N$

Important Questions on Rotational Mechanics

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