System of Particles and Rotational Motion

Physics Part I

Physics

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As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (TakeÂ *gÂ *= 9.8 m/s^{2})

(Hint: Consider the equilibrium of each side of the ladder separately.)

The given question is illustrated in the figure below:Â *N*_{B}Â = Force exerted on the ladder by the floor point BÂ *N*_{C}Â = Force exerted on the ladder by the floor point CÂ *TÂ *= Tension in the ropeÂ

BA = CA = 1.6 mÂ

DE = 0. 5 mÂ

BF = 1.2 mÂ

Mass of the weight,Â *m*Â = 40 kgÂ

Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.Â

Î”ABI and Î”AIC are similar triangles.Â

âˆ´ Â Â Â BI = ICÂ

Hence, I is the mid-point of BC.Â

DE || BCÂ

Â BC = 2 Ã— DE = 1 mÂ

AF = BA â€“ BF = 0.4 m Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦Â **(i)Â **

D is the mid-point of AB. Â

Hence, we can write,Â

AD = Â Ã—Â BAÂ =Â 0.8 m Â Â Â Â Â Â Â Â Â Â ...

Using equationsÂ

FE = 0.4 mÂ

Hence, F is the mid-point of AD.Â

FG||DH and F is the mid-point of AD.

Hence, G will also be the mid-point of AH.Â

Î”AFG and Î”ADH are similarÂ

Â Â Â Â

In Î”ADH,Â

AH = (AD

Â Â Â = (0.8

Â Â Â =Â 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force.Â

For rotational equilibrium of the ladder, the net moment about A isÂ

-N

-N

(N

N

Adding equationsÂ

N

N

For rotational equilibrium of the side AB, consider the moment about A,Â

Â Â Â Â Â Â Â -N

-245Â Ã—Â 0.5 + 40 X 9.8Â Ã—Â 0.125 + TÂ Ã—Â 0.76Â =Â 0 Â

âˆ´ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â T = 96.7 N.

T is the required tension in the rope.Â

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Define centre of mass.

Centre of mass of a body or a system of bodies is a point at which the entire mass of the body or system is supposed to be concentrated.Â

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Is it necessary that there should be matter at the centre of mass of system?

No, it is not necessary that there be matter at the centre of mass of the system.

For e.g., if two equal point masses are separated by certain distance, the centre of mass lies at the mid point of two point masses and there is no mass at that point.

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What is the significance of defining the center of mass of a system?

The motion of n particle system can be reduced to one particle motion.

An equivalent single point object would enable us to discuss the overall motion of the system.Â

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What is the need of centre of mass?

Newtonâ€™s second law of motion is strictly applicable to point masses only. To apply the Newton's law of motion to rigid bodies, the concept of centre of mass is introduced.

The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object.Â

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Is it necessary for centre of mass to lie within the body?

No, centre of mass needs not to lie within the body. It is not necessary that the total mass of the system be actually present at the centre.

The position of the centre of mass is calculated using the usual Newtonian type of equations of motion.Â

The position of the centre of mass is calculated using the usual Newtonian type of equations of motion.Â

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