Fantastic Physics: Heroic Deed in Argentina!

 Fantastic Physics Explains a Heroic Deed in Argentina

By: Rachel Valdivieso

Topic: Forces and Work

http://news.bbc.co.uk/2/hi/8509776.stm

In February 2010, BBC news reported that an Argentine man saved a van, which was parked in the middle of the train tracks, from being crushed by an oncoming train. The man was able to quickly jump off his friend’s motorcycle and push the van from the tracks and avoid the oncoming train within seconds.  This is an example of fantastic physics, because he must have applied a very large magnitude of force to push the van from the train tracks.

In appreciation of his courageous deed, his applied force on the van will be calculated:

Assuming the net-work done on the 2250 kg van is 9.58  10⁷ J and its displacement is 178.87 ft. and the coefficient of kinetic friction (μ_k) is 0.4, the force the man applied can be calculated. (Assume g=9.8m/s² )

The first step in solving this problem is to convert the units: 178.87 ft. must be converted to meters. Given the unit conversion 1ft.=0.3408m:

178.87ft. ×.3408m/ft.=60.96m=the van’s displacement (d)

All the forces acting on the van must be considered: F_G, F_N, F_App, and F_Fr

F_G, the force due to gravity, does not account for any work being done on the van, because it acts downwards in the y-direction, which is perpendicular to the van’s displacement in the x-direction. 

F_G×d×Cos(θ)=(2250kg)(9.8m/s²)(60.96m)[cos(90°)]=OJ=W_G, since cos(90°)=0.

F_N, the normal force that the ground exerts upwards on the van, also does not account for any work being done on the van, because it acts upwards in the y-direction, perpendicular to the van’s displacement in the x-direction. Since the normal force is perpendicular to the van’s displacement, θ=90°, so multiplying cos(90°) by the F_N and d gives 0J of work done by the normal force.

F_App, the force applied by the man on the van, does account for a portion of the net-work being done on the van, because it acts parallel to the van’s displacement in the x-direction.

F_App×d×cos(0°)=W_App

F_Fr, the force of friction due to the van’s contact with the ground, also accounts for a portion of the net-work being done on the van, because it acts parallel to the van’s displacement in the x-direction.

F_Fr×d×cos(180°)=W_Fr

Summing up the Forces in the y-direction and setting them equal to zero, since the van does not accelerate in the y-direction, a_y=0.

ΣF_y=ma_y=0=F_N-F_G, so F_N=F_G

W_Net=W_App+W_Fr

W_Fr=[cos(180°)]×(F_Fr)×(d), since F_Fr=F_Nμ and F_N=F_G=mg, so F_Fr=μmg;

W_Fr=[cos(180°)]×μ × m × g × d

W_Fr= (-1)×(0.4)×(2250kg)×(9.8m/s²)×(60.96m)≈-5.3767×10⁵J

W_App=F_App×d×cos(0°)=F_App×60.96m×1=60.96F_App

Summing up F_App and F_Fr:

9.58×10⁷J=60.96F_App-5.3767×10⁵J

Solve for F_App:

F_App=1.58×10⁶N

The answer should be given to one significant figure:

F_App≈2×10⁶N