Question

This is a car with 1500 kg mass
The centre of mass is 60 m away
The distance between 2 wheels is 100m. Find the force acting on both the wheels​

Answer 1

Answer:

Given:

Mass of car = 1500 kg

Centre of mass = 60 m (let's assume it is 60m behind the front wheel)

Distance between the wheels = 100m

To find:

Force acting on the wheels.

Diagram:

Please refer to the attached photo to understand better.

Concept:

In these type of questions, we need to apply "Equilibrium of translation" and "Equilibrium of rotation"

∑(MF) = 0 and ∑(Mζ) = 0 ,

where MF is Moment due to force (translational equilibrium) and Mζ is the moment due to Torque (rotational equilibrium)

Calculation:

Let Force on Front wheel = N1

Force on Back wheel = N2

So as per translational equilibrium ,

∑(MF) = 0

=> N1 + N2 - mg = 0 ............(1)

Now,

As per rotational equilibrium with respect to Point "A" ( refer to the diagram)

∑(Mζ) = 0

=> (N1 ×0) - (N2 × 100) + {(mg) × 60} = 0

=> (N2 × 100) = 60 × mg

=> N2 = (60 × m × 10)/100

=> N2 = 6m

=> N2 = 6 × 1500 = 9000 Newton

=> N2 = 9000 N

Putting value of N2 in eq.(1)

=> N1 + 9000 - mg = 0

=> N1 + 9000 = mg = 1500 × 10

=> N1 = 15000 - 9000

=> N1 = 6000N.

So Force on front wheel = 6000 N.

And Force on Back wheel = 9000 N

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of car = 1500 kg
• Center of mass Distance = 60 meters.

$\huge{\bold{\underline{Explanation:-}}}$

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Applying the Principle of moments,

The Net Force Applied will be Zero.

and,

The Net Torque acting on the System will also be zero.

Assumption:-

Taking Centre of mass Distance from the wheel "P" I.e. Front wheel. & Taking wheel "P" as the axis of rotation.

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Now,

⇒ Σ F = 0.

(Σ F represents Summation of all Forces)

⇒ N₁ + N₂ - mg = 0

⇒ N₁ + N₂ = mg -----(1)

Note:-

Here we took Weight as negative as it is acting opposite to the Normal reaction.

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⇒ Σ τ = 0.

(Σ τ represents Summation of all Forces)

⇒ τ₁ + τ₂ - τʷ = 0

(τʷ = Torque due to weight)

Taking Axis of rotation as "P"

⇒ (N₁ × 0) - (N₂ × 100) + (mg × 60) = 0

⇒ 0 - N₂ × 100 = - 60 mg

⇒ N₂ = 60mg/100

⇒ N₂ = 6 mg/10

⇒ N₂ = 6 × m × 10/10

[g = 10 m/s²]

⇒ N₂ = 6m

Substituting the value of Mass

⇒ N₂ = 6 × 1500

⇒ N₂ = 9000 N.

(Here we took Negative Torque for N₂ as it is Antagonistic to the rest two Torques)

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Substituting the value in Equation (1).

⇒ N₁ + N₂ = mg -----(1)

⇒ N₁ + 9000 = 1500 × 10

⇒ N₁ + 9000 = 15000

⇒ N₁ = 15000 - 9000

⇒ N₁ = 6000 N.

So, The Normal reactions at Wheel "P" (First wheel) is (N₁) 6000 N. & The Normal reactions at Wheel "Q" (Second wheel) is (N₂) 9000 N.

#refer the attachment for figure.

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