. For the arrangement shown in the figure the tension in the string is
A: 6 N
B: 6.4 N
C: 0.4 N
D: Zero
Best Answer
Explanation:
The frictional constant
u=0.8
∝=tan-1(0.8)
⇒∝=39º
As the inclined plane angle 37º<39º.So the block will not slight down
So force applied in the block Fapplied=mgsin(37º)
Now the equilibrium of force can be written as
T+Fapplied=mgsin(37º)
⇒T=0
Final Answer:
The tension in the arrangement is T=0.Hence option (D) is correct.
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