Question 8
A small ball of weight 10 N is suspended by two cords A and B as shown in the
figure. Values of tensions in the strings A and B are
37°
53°
A
B
(1)
(2)
(3)
(4)
80 N and 60 N respectively.
6 N and 4 N respectively
6 N and 8 N respectively
8 N and 6 N respectively.
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Answers
Answer:
2
Explanation:
the correct answer is 2nd
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Answer:
The values of tensions in strings A and B are 6N and 8N respectively.
Explanation:
Given:
Weight of ball (W) = 10N
Angle made by string A = 37°
Angle made by string B = 53°
To find:
Tension in strings A and B (T1 and T2)
Step 1:
Taking resolution of forces in the horizontal direction,
T1 cos37° = T2 cos53°
0.799×T1 = 0.602×T2
T1 = (0.602/0.799) T2
T1 = 0.753×T2 ...(i)
Step 2:
Taking resolution of forces in the vertical direction,
T1×sin37° + T2×sin53° = 10
0.602×T1 + 0.799×T2 = 10
Substituting the value of T1, we get
0.753×0.602×T2 + 0.799×T2 = 10
0.453×T2 + 0.799×T2 = 10
1.252×T2 = 10
T2 = 10/1.252
T2 = 7.98
T2 ≈ 8N
Step 3:
Substituting the value of T2 in (i),
T1 = 0.753×T2
T1 = 0.753×8
T1 = 6.024
T1 ≈ 6N
Therefore, the tension in string A is 8N and in string B is 6N.
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