1
11. Blocks A and B in the figure are connected by a bar of negligible weight. Mass of each block is
170Kg and a = 0.2 and Hp = 0.4 ,where , and I are the coefficients of limiting fricion
between blocks and plane. The force developed in the block is given by (g = 10 ms?)
[ a
a) 150 N
В.
b) 75 N
c) 200 N
8
d) 250 N
15
Answers
Answered by
9
Explanation:
ANSWER
cosθ=
17
5
sinθ=
17
8
f
A
and f
B
be the frictional force action of block A and B respectively.
Acceleration of blocks is a
Then for block A;
mgsinθ−f
A
−T=ma
⇒170×10×
17
8
−μ
A
mgcosθ−T=ma
⇒10×10×8−0.2×170×10×
17
15
−T=170a
⇒500−T=170a ..............(1)
Now, consider block B;
T+mgsinθ−f
B
=ma
⇒T+170×10×
17
8
−μ
B
mgcosθ=ma
⇒T+800−0.4×170×10×
17
15
=170a
⇒T+800−600=170a
⇒T+200=170a ..............(2)
From eq. (1) and (2);
500−T=T+200
⇒2T=300
⇒T=150N
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