Test Index

Systems of Particles and Rotational Motion

© examsnet.com
Question : 8
Total: 33
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Solution:  
Let AB be the uniform bar of weight W suspended at rest by the two strings OA and O′B which make angles 36.9° and 53.1° respectively with the vertical.
∴∠OAA′=90°–36.9°=53.1°
Similarly ∠O′BB′=36.9°
AB=2m,AC=dm.
Let T1‌and‌T2 be the tensions in the strings OA and O′B respectively and their rectangular components are shown in the figure.1As the rod is at rest, so the vector sum of the forces acting along A′B′ axis and ⊥ to it are zero i.e.
–T1‌cos‌53.1°+T2‌cos‌36.9°=0...(i)
and T1‌sin‌53.1°+T2‌sin‌36.9°‌–‌W=0...(ii)
Taking the torques about A and equating the sum of torques to zero, we get
–(T2‌sin‌36.9°)×2+Wd=0
or T2=
Wd
2‌sin‌36.9°
...(iii)
∴ From (ii) and (iii), we get
T1‌sin‌53.1°=W−T2‌sin‌36.9°=W−
Wd
2
∴T1=
W
sin‌53.1°
 (1−
d
2
)...(iv)
∴ From (i), (iii) and (iv), we get T1‌cos‌53.1°=T2‌cos‌36.9°
or
W (1−
d
2
)
sin‌53.1°
‌cos‌53.1°=
Wd‌cos‌36.9°
2‌sin‌36.9°
or
W(1−
d
2
)
tan‌53.1°
=
Wd
2‌tan‌36.9°

or
(1−
d
2
)
1.3319
=
d
2
0.7508

or 1−
d
2
=
d
2
×
1.3319
0.7508

=
d
2
×1.7740=0.8870d
or 0.5d+0.8870d=1 or d=
1
1.3870
=0.721 m=72.1 cm
© examsnet.com
Go to Question:
Prev Question
Next Question