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Vault Scene Clip:

Starts at 1:15

Ends at 1:55

Our clip is from the movie Fast Five, where Vin Diesel and Paul Walker are attempting to steal a safe filled with $1 million by attaching it to the back of two Dodge Chargers. In the clip, the cars are accelerating for a period with no movement of the safe until eventually the safe starts to move and gain momentum.

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Givens

Coefficients of friction:

rubber on concrete¹ = 0.8
steel on concrete² = 0.45

According to the movie, the cars they are driving are Dodge Charger STR-8’s, which weigh 1900 kg³, so mass_{2 cars}= 3800 kg

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Givens (continued)

According to the movie, the vault is “ten tons of top-of-the-line security” → 10 tons = 9100 kg
It is mentioned that the vault contains 49 kg bags of $1 million each. The vault contains $100 million total.

49 kg * 100 packs = 4900 kg of money

Mass_{vault + money}= 9100 + 4900 = 14000 kg

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In this snapshot the safe is dragging across the floor at high speeds

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Finding v₂ and a_1-2

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Newton’s Third Law

Newton’s third law states that: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the forces on the second object.

This applies in several parts of our movie scene.
First is the force the cars use to try and move the vault. This force generating from the car is essentially equal to the static friction that is keeping the vault from breaking free of its movements in the beginning.
The second application of the third law can be seen with the tension in the cord that is attached to both the back of the car and the vaults. This tension is set from the equal force of the cars force pulling on the vaults end and the vaults resistance/friction pulling on the rear end of the cars.

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Force Diagrams!

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Finding Friction on the Vault

G: μ for steel on concrete = 0.45, Fn_vault = 140000 N

U: F_fr

I: F_fr = μ * Fn

D: 0.45 * 140000 = 63000 N

E: F_fr = 63000 N

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Updated Diagrams

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Finding Tension on the Vault

G: a = 0.97 m/s², m_vault = 14000 kg, F_fr = 63000 N

U: F_Tension

I: F_net = m * a, F_Tension = F_net + F_fr

D: F_Tension = (14000)(0.97) + 63000 = 76580 N

E: F_Tension = 76580 N

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Updated Diagrams

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Finding the Force of the Cars

G: a = 0.97 m/s², m_cars = 3800 kg, F_Tension = 76580 N

U: F_Cars

I: F_net = ma, F_Cars = F_net + F_Tension

D: F_Cars = (3800)(0.97) + 76580 = 80266 N

E: The force of the cars is 80266 N

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Final Diagrams

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What does this mean?

In order for the cars to generate a friction force of 80266 N, the μ value between the rubber wheels and the car would have to be very high.

G: F_Cars = 80266 N, F_N = 38000 N

U: μ

I: F_fr = μ * F_N

D: 80266 = μ * 38000, μ = 2.11

E: A μ value of 2.11 is necessary for the cars to be able to tow and accelerate the vault at the speeds witnessed in the film.

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Conclusion

In order for the cars to pull the vault as fast as they did in the scene, there would need to be a friction coefficient of 2.11 acting on the wheels
This is highly improbable to begin with for any two substances, and it also disagrees with the value we found online of 0.8 for the coefficient between concrete and rubber.
This scene is cool, but not feasible.
Go physics!

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