The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal level. A body of mass 800g is hung on the string at a point O, 2.4m from Y. If the system is kept in equilibrium by a horizontal force P acting on the body and the tensions are equal, calculate:
(a) < XOY;
(b) the magnitude of the force P;
(c) the tension T in the string.
(a) < XOY;
(b) the magnitude of the force P;
(c) the tension T in the string.
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Correct Answer: Option
Explanation:
Cos \(\theta\) = \(\frac{(2.4)^2 + (1.8)^2 - (3)^2}{2(2.4)(1.8)}\)
cos \(\theta\) = \(\frac{5.76 + 3.4 - 9}{8.64}\)
cos \(\theta\) = \(\frac{0.0}{8.64}\) = 0
\(\theta\) cos\(^{-1}\) 0 = 90\(^o\)
< XOY = 90\(^o\)
(b) Magnitude of the force p
cos \(\alpha\) = \(\frac{(3)^2 + (2.4)^2 - (1.8(^2}{2(3)(2.5)}\)
cos \(\alpha\) = = \(\frac{9 + 5.76 - 3.24}{14.4}\)
cos \(\alpha\) = \(\frac{11.52}{14.4}\)
cos \(\alpha\) = 0.8
\(\alpha\) = cos \(^{-1}\) 0.8
= 36.86\(^o\)
\(\frac{p}{2.4} = cos 36.86^o\)
p = 2.4 x 0.8
p = 1.92N
(c) The tension T in the string
T\(^2\) = 8\(^2\) + 1.92\(^2\)
T\(^2\) = 64 + 3.59
Tv = 67.69
T = \(\sqrt{67.69}\)
T = 8.23N
Cos \(\theta\) = \(\frac{(2.4)^2 + (1.8)^2 - (3)^2}{2(2.4)(1.8)}\)
cos \(\theta\) = \(\frac{5.76 + 3.4 - 9}{8.64}\)
cos \(\theta\) = \(\frac{0.0}{8.64}\) = 0
\(\theta\) cos\(^{-1}\) 0 = 90\(^o\)
< XOY = 90\(^o\)
(b) Magnitude of the force p
cos \(\alpha\) = \(\frac{(3)^2 + (2.4)^2 - (1.8(^2}{2(3)(2.5)}\)
cos \(\alpha\) = = \(\frac{9 + 5.76 - 3.24}{14.4}\)
cos \(\alpha\) = \(\frac{11.52}{14.4}\)
cos \(\alpha\) = 0.8
\(\alpha\) = cos \(^{-1}\) 0.8
= 36.86\(^o\)
\(\frac{p}{2.4} = cos 36.86^o\)
p = 2.4 x 0.8
p = 1.92N
(c) The tension T in the string
T\(^2\) = 8\(^2\) + 1.92\(^2\)
T\(^2\) = 64 + 3.59
Tv = 67.69
T = \(\sqrt{67.69}\)
T = 8.23N