posted by Sandhya on .
Three blocks of mass 1kg,2kg and 3kg move on a frictionless surface and a horizontal force 46 N acts on the 3kg block.1kg and 2kg blocks are in contact with each other while 1kg and 3kg blocks are connected by a chord.
(a) Determine the acceleration of the system
(b) Determine the tension in the cord connecting the 3.0 kg and the 1.0 kg blocks.
(c) Determine the magnitude of the force exerted by the 1.0 kg block on the 2.0 kg block.
a) The applied force of F = 46N acts on the system of all three blocks. You can ignore the internal foces (cord and contact) between blocks and apply F = (M1 + M2 + M3) a, where M1 = 1 kg, M2 = 2 kg and M3 = 3 kg.
a = 46 N/6 kg = 23/3 m/s^2
b) Let f1 be the cord tension. Only F and f1 act on M3, the 3 kg mass. Apply Newton's law.
F - f1 = M3 * a
Solve for f1. Use the value of a from part (a)
(c) The only force applied to M2 is the contact force from M1. Call it f2. Solve
f2 = M2 a
Thanks a lot for your timely help.