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Engineering Entrance Exams

A body of 2 kg is fastened to a fixed point by means of a string of length 50 cm.
It is acted upon a horizontal force F and rest at a distance of 40 cm from the vertical
line through the fixed point. Find the value of F and tension in the string � riB� HT�UHTline-height: normal;mso-layout-grid-align:none;text-autospace:none'>the floor and watt .

Profile image of RAJESH  RAVAL
12 Years agoGrade
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1 Answer

Profile image of Sumit Majumdar
11 Years ago
Dear student,
Using the FBD for the figure, we find that:
Tcos\theta=mg; Tsin\theta=F
Hence,
T=\frac{mg}{cos\theta }=\frac{2\times 10\times 5}{3}=\frac{100}{3} N
and so,
F=\frac{100}{3}\times \frac{4}{5}=\frac{80}{3} N
Regards
Sumit


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