Stick in water

Classical Mechanics Level 3

A rod of length 6m has specific gravity 25/36 . One end of rod is tied to 5m long rope, which in turn is tied to the floor of a pool 10m deep as shown .Find the length of part of the rod,which is out of water.


The answer is 1.

Manish Ssaid by Manish ßaid , 26, India

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2 solutions

Lynn Saibor
May 4, 2014

W(wood) = (25/36)(9.81)(area 6 m); F(buoyant)= 9.81(area L(submerged length))

NOTE: We could cancel area here so we could have - W(wood)=40.875 N & F(buoyant)=9.81 L

Let θ be the angle of inclination of the rod. Therefore; W=3 sin θ & F= (L/2) sin θ

∑M @ A = 0

(40.875)( 3 sin θ) = (9.81 L)(L/2) sin θ 25 = L^2 5=L, so 5 m of the rod length is submerged hence, the answer is 1 m.

9 Helpful 0 Interesting 2 Brilliant 5 Confused

I think the torque produced on the rod should be mg×3cos(theta) because we should take the displacement vector in the perpendicular direction of the force applied

Shivani Lohiya - 2 years, 3 months ago

And also it should be F=(9.81L)(L/2)cos(theta) am I wrong

Shivani Lohiya - 2 years, 3 months ago
Jeremiah Jocson
Mar 5, 2015

the rod length is 6 m, since the sp gravity is given 25/36 we can easily calculate the length of the rod which is submerge "(6)(25/36) = 4.6167 m" . Since the minimum length is 5 m, therefore the answer is "6 m - 5 m = 1 m"

3 Helpful 0 Interesting 1 Brilliant 5 Confused

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