A rigid rod XY has negligible mass and length 75 cm. The rod is suspended from a fixed point Pby a string attached to end X. An object of mass 11 kg is suspended by a string that is attached tothe rod at a distance of 25 cm from end X, as shown.75 cm25 cm11 kgPX YWhich vertically upward force acting on end Y of the rod would hold the rod horizontally inequilibrium?A 3.7 N B 33 N C 36 N D 320 N
Question
A rigid rod XY has negligible mass and length 75 cm. The rod is suspended from a fixed point Pby a string attached to end X. An object of mass 11 kg is suspended by a string that is attached tothe rod at a distance of 25 cm from end X, as shown.75 cm25 cm11 kgPX YWhich vertically upward force acting on end Y of the rod would hold the rod horizontally inequilibrium?A 3.7 N B 33 N C 36 N D 320 N
Solution
To solve this problem, we need to understand that the rod is in equilibrium, which means the sum of the torques (or moments) about any point is zero. We can choose point X as the pivot point because we know the distances and forces acting on the rod.
The weight of the 11 kg object acts downwards and creates a clockwise moment about point X. This moment is given by the force due to gravity (mass x gravity) times the distance from the pivot point.
So, the moment due to the 11 kg mass is 11 kg * 9.8 m/s^2 * 0.25 m = 26.95 Nm.
The force at Y must create an equal and opposite moment to balance this, so it must create a counterclockwise moment of 26.95 Nm. The distance from Y to the pivot point X is 0.75 m, so we can find the force at Y by rearranging the moment equation:
Force = Moment / distance Force = 26.95 Nm / 0.75 m = 35.93 N, which is approximately 36 N.
So, the answer is C, 36 N.
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