In April 1974, John Massis of Belgium managed to move two passenger railroad car
ID: 1593344 • Letter: I
Question
In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed 700 kN (about 71 tons). Assume that he pulled with a constant force that was 2.5 times his body weight, at an upward angle of 26 from the horizontal. His mass was 78 kg, and he moved the cars by 3 m. Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.
Explanation / Answer
F = 2.5*(Body Weight) = 2.5*(m*g) = 2.5*(78*9.8) = 1911 N
x-directional force on the cars,
F(x) = F cos = F cos26 = 1911 cos26 = 1717.6 N
So, x-directional acceleration on the cars,
a(x) = F(x) / m = 1717.6 / (700*10^3/g) = 1717.6*9.8 / 700*10^3 = 0.024 m/s^2
Now for the speed of the cars at the end of the pull, we can use the below formula,
v^2 = u^2 + 2*a*s
where, u = initial speed = 0, a = a(x) and s = 3 m (given in the question)
=> v^2 = 0 + 2*0.024*3 = 0.144 => v = sqrt(0.144) = 0.379 m/s