## Tuesday, November 6, 2012

### Fantastic Physics: Heroic Deed in Argentina!

Fantastic Physics Explains a Heroic Deed in Argentina

By: Rachel Valdivieso

Topic: Forces and Work

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 (μk) 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: FG, FN, FApp, and FFr

FG, 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.

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

FN, 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 FN and d gives 0J of work done by the normal force.

FApp, 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.

FApp×d×cos(0°)=WApp

FFr, 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.

FFr×d×cos(180°)=WFr

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

ΣFy=may=0=FN-FG, so FN=FG

WNet=WApp+WFr

WFr=[cos(180°)]×(FFr)×(d), since FFr=FNμ and FN=FG=mg, so FFr=μmg;

WFr=[cos(180°)]×μ × m × g × d

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

WApp=FApp×d×cos(0°)=FApp×60.96m×1=60.96FApp

Summing up FApp and FFr:

9.58×107J=60.96FApp-5.3767×105J

Solve for FApp:

FApp=1.58×106N

The answer should be given to one significant figure:

FApp≈2×106N