oo Verizon LTE 5:08 PM * 1596D net Three blocks are in contact with one another
ID: 1772518 • Letter: O
Question
oo Verizon LTE 5:08 PM * 1596D net Three blocks are in contact with one another on a frictionless horizontal surface as in the figure below. A horizontal force r is applied to m1 where m1 = 2.15 kg, m2 = 3.73 kg, m3 = 4.92 kg, and F = 19.2 N. (a) Find the acceleration of the blocks. m/s (b) Find the net force on each block. (c) Find the magnitudes of the contact forces between the blocks. Need Help? Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F (Fig, P5.45). Suppose that F = 69.0 N, m1 = 13.0 kg, m2 = 17.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.102. Figure P5.45 (b) Determine the tension T Determine the magnitude of the acceleration of the system. m/s2 Two blocks, each of mass m = 3.20 kg are hung from the ceiling of an elevator as in the figure belowExplanation / Answer
given three masses
m1 = 2.15 kg
m2 = 3.73 kg
m3 = 4.92 kg
F = 19.2 N
alos, the track is frictionless
a. let acceleration of the blocks be a
then
From newtons second law
(m1 + m2 + m3)*a = F
(2.15 + 3.73 + 4.92)a = 19.2
a = 1.7787 m/s/s
b. net force on each block can be found using net acceleration of each block
net force on block 1 = m1*a = 2.15*a = 3.82421 N
net force on block 2 = m2*a = 3.73*a = 6.63456 N
net force on block 3 = m3*a = 4.92*a = 8.75122 N
c. let force between ma and m2 be C1, and that between m2 and m3 be C2
then from applygin force balance on the first block
F - C1 = m1*a
C1 = 19.2 - 2.15*1.7787 = 15.375795 N
similiarly from the third block
C2 = m3*a = 4.92*1.7787 = 8/75122 N
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