An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1 = 584 kg, m2 = 746 kg and m3 = 344 kg have blocked a busy road. The rocks are side by side blocking the road and lined up from left to right in order as m1, m2 and m3. The city calls a local contractor to use a bulldozer to clear the road. The bulldozer applies a constant force to m1 to slide the rocks off the road. Assuming the road is a flat frictionless surface and the rocks are all in contact, what force, FA, must be applied to m1 to slowly accelerate the group of rocks from the road at 0.400 m/s2?Use the value found above for FA to find the force, F12, exerted by the first rock of mass 584 kg on the middle rock of 746 kg
The total mass to be moved = 584 + 746 + 344 = 1674 kg. Force applied to all the blocks (FA) = ma = 1674 * 0.400 = 669.6 Newtons. The force applied to just the first rock - ma = 584 * 0.400 = 233.6 Newtons. So the remaining force applied to the last two rocks (F12) = 669.6 - 233.6 = 436 Newtons.
Clearly, the force applied to the first rock is FA = Σm*a = 1674*.400 = 669.6 N. F12 = a*(m2 + m3) = 436 N F23 = a*m3 = 137.6 N This may be a trick question; if the road is frictionless, the bulldozer won't be able to apply a force to anything.....!
