A system of two blocks and a light string are kept on two inclined faces(rough) as shown. All the req data are mentioned in diagram.pulley is light and frictionless.If the system is released from rest then what is the range of the tension in the string?
Asked by m.nilu
16th August 2018,
7:48 PM
Answered by Expert
Answer:
(Peculiar thing about the right triangle that has other angles 37º and 53°
is the sides are in the ratio 3:4:5, this is used here to get sine and cosine values)
From the F.B.D of 10 kg block, we write Newton's law as,
10×g×sin37-T-μl×10×g×cos37 = 10×a ..........................(1)
From the F.B.D of 5 kg block, we write Newton's law as,
T - 5×g×sin53-T-μr×5×g×cos53 = 5×a ..........................(2)
By adding (1) and (2), we get { 10×sin37 - 5×sin53}g - {μl×10×cos37+μr×5×cos53}g = 15×a ...........(3)
In eqn.(3), first two terms are force due to weight of blocks on inclined plane surfaces,
third and forth term are due to frictional force.
To get acceleration, first bracketed term should exceed the frictional force term.
Let us calculate each term separately
{ 10×sin37 - 5×sin53}g= 2g ...................(4)
{μl×10×cos37+μr×5×cos53}g = {0.3×10×0.8 + 0.1×5×0.6}g = 2.7g ............(5)
from (4) and (5),we see that,sum of frictional forces exceed thesum of force due to weight of the blocks.
Hence the blocks are stationary and acceleration is zero.
Tension force in the string is calculated from eqn.(2) using a = 0 as follows
T = 5×g×sin53-T + μr×5×g×cos53 = 5×9.8×0.8+0.1×5×9.8×0.6 = 42.14 N
Answered by Expert
17th August 2018,
7:20 AM
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