James Gomer constructs a uniform rod with a weight of 40.0 N and a length of 1.0

Question

James Gomer constructs a uniform rod with a weight of 40.0 N and a length of 1.00 m. It is hinged to a wall (at the left end), and held in a horizontal position by a vertical string (at the right end), as shown in the figure below. What is the magnitude of the force exerted by the string? Do not forget the WEIGHT of the rod at the CG. 40.0 N 10.2 N 5.20 N 20.0 N 30.m.N Above we had James Gomer holding a uniform 40 N rod of length 1 m horizontally with a hinge on its left end and a vertical string on the right end. Now suppose a fully laden swallow with a mass of 2.00 kg lands and squawks loudly on the rod at a point 30.0 cm to the right of the hinge. Now what is the force exerted by the string to keep the rod horizontal? 24.8 N 25.9 N 30.5 N 19.2 N

Explanation / Answer

Part A.

Weight of rod = mg = 40 N

Length of rod = 1 m

Weight acts at center of gravity which is at midpoint of the rod.

Now balancing the torque about hinge.

40×(1/2) = T × (1)

Or. T = 20 N

Hence force by string = 20 N

Part B.

Weight of swallo = 2×9.8 = 19.6 N

Distance of swallo from hinge = 30cm = 0.30 m

Againg balancec the torque about the hinge

40(0.5) + 19.6×(0.30) = T(1)

Or. T = 25.9. N

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