Physics

Mass of the car, m = 1800 kg

Distance between the front and rear axles, d = 1.8 m

Distance between C.G. and the front axle = 1.05 m

Let Rf and Rb be the force exerted from ground at front and rear axles respectively.

Rf + Rb = mg = 1800 x 9.8 N = 17640 N ……. (i)

For rotational equilibrium around C.G. we have Rf x 1.05 = Rb x (1.8 – 1.05)

R_f x 1.05 = R_b x 0.75

R_f/R_b = 0.75 / 1.05

R_f = 0.71 R_b ……. (ii)

From equation (i), we get

0.71 R_b + R_b = 17640

R_b = 10316 N

R_f = 7324 N

Therefore, force exerted on each front wheel = R_f/2 =7324/2 = 3662 N

Force exerted on each rear wheel, R_b/2 = 10316 / 2= 5158 N

A free body diagram needs to be drawn.

The length of the bar, l = 2 m

T1 and T2

At translational equilibrium, we have $\stackrel{?}{L_q}=\stackrel{?}{L_R}$ = $T_1$

(T1 / T2) = ( $T_2$ / $T_1\sin ?36.9°$ = 4/3

T1 = (4/3)T2

For rotational equilibrium, on taking the torque about the centre of gravity, we have

T1 $T_2\sin ?53.1°$ x d = T2 $\sin ?53.1°$  (2-d)

T1 x 0.8d = T2 x 0.6 (2-d)

(4/3)T2 x 0.8d = T2 x 0.6 (2-d)

(4/3) x 0.8d = 0.6 (2-d)

1.07d = 1.2 – 0.6d

d = 0.72

So the c.g. of the given bar lies at 0.72 m from its left end.

Let at certain instant two particles be at points P and Q, as shown in the figure.

Angular momentum of the system about point P

$\stackrel{?}{L_p}$ = mv x 0 + mv x d = mvd ……. (i)

Angular momentum of the system about point Q

$\stackrel{?}{L_q}$ = mv x d + mv x 0 = mvd ……. (ii)

Consider a point R, which is at a distance y from point Q such that QR = y

PR = d – y

Angular momentum of the system about point R

$\stackrel{?}{L_R}$  = mv x (d – y) + mv x y = mvd – mvy + mvy = mvd ……. (iii)

Comparing equations (i), (ii) and (iii) we get

$\stackrel{?}{L_p}$ = $\stackrel{?}{L_q}$ = $\stackrel{?}{L_R}$ …… (iv)

Let $\stackrel{?}{c}$ = $\stackrel{?}{b}$ ,  $\stackrel{?}{a}$ = $\stackrel{?}{a}$ ,  $\stackrel{?}{OA}$ = $\stackrel{?}{b}$

Let $\stackrel{?}{OB}$ be a unit vector perpendicular to both b and c. Hence $\stackrel{?}{c}$ and a have the same direction

Now $\stackrel{?}{OC}$ = bc $\stackrel{?}{n}$ $\stackrel{?}{n}$

= bc $\stackrel{?}{b}*\stackrel{?}{c}$ $\sin ?\theta$ = bc $\stackrel{?}{n}$

Now $\sin ?90°$ )= a. (bc $\stackrel{?}{n}$ ) = abccos $\stackrel{?}{n}$ ? = abccos0° = abc = Volume of the parallelepiped

Let AB is equal to the vector a and AC be equal to the vector b.

Consider two vectors $\sin ?\theta$ = $\frac{CN}{AC}$ = $\frac{CN}{\left|\stackrel{?}{B}\right|}$

$\stackrel{?}{AC}$ = $\stackrel{?}{b}$ inclined at an angle $\theta$

MN = $\stackrel{?}{b}\sin ?\theta$

| $\stackrel{?}{a}$ $*\stackrel{?}{b}$ | = | $\stackrel{?}{a}$ | | $\stackrel{?}{b}|\sin ?\theta =AB*AC$

The area of ΔABC, we can write the relation

Area of Δ ABC = $\frac{1}{2}$ AB $*AC$ = $\frac{1}{2}\left. [\stackrel{?}{a}*\stackrel{?}{b}\right]$

The child is sitting on the trolley and there is no external force, hence it is a single system. The velocity of the centre of mass will not change, irrespective of any internal motion.

If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have, : (my+35.5mx)/ (m+35.5m) = 0

 mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

All the structures specified are symmetric bodies with uniform mass density. For all these bodies, their centre of mass will lie in their geometric centres.

Not necessarily, the centre of gravity of a circular ring is at the imaginary centre of the ring.