1) As shown in the figure, a 10.0 m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle \(\theta = 48^\circ\) with the bar. The bar is uniform and weighs 81.4 N. What distance x from the hinge should a 10.0 kg mass be suspended for the tension in the rope to be 177 N?

Name: 1) As shown in the figure, a 10.0 m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle θ = 48^∘ with the bar. The bar is uniform and weighs 81.4 N. What distance x from the hinge should a 10.0 kg mass be suspended for the tension in the rope to be 177 N?
Uploaded: 2023-08-01T04:15:06-08:00
Duration: 2 min 30 s
Channel: Prabhu Ramji
Description: 1) As shown in the figure, a 10.0 m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle θ = 48^∘ with the bar. The bar is uniform and weighs 81.4 N. What distance x from the hinge should a 10.0 kg mass be suspended for the tension in the rope to be 177 N?

          1) As shown in the figure, a 10.0 m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle \(\theta = 48^\circ\) with the bar. The bar is uniform and weighs 81.4 N. What distance x from the hinge should a 10.0 kg mass be suspended for the tension in the rope to be 177 N?

1) As shown in the figure, a 10.0 m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle θ = 48^∘ with the bar. The bar is uniform and weighs 81.4 N. What distance x from the hinge should a 10.0 kg mass be suspended for the tension in the rope to be 177 N?

Added by Juan Manuel S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prabhu Ramji

08/01/2023

Step 1

The weight of the bar can be considered to act at its center of mass, which is at a distance of 5.0m from the pivot point. The torque produced by the weight of the bar is given by τ = F * r, where F is the weight of the bar (81.4N) and r is the distance from the Show more…

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pivot point. What is the magnitude of the torque about the pivot point produced by this force? As shown in the figure, a 10.0m long bar is attached by a frictionless hinge to a wall and held horizontal by a light rope that makes an angle θ = 48° with the bar. The bar is uniform and weighs 81.4N. What distance x from the hinge should a 10.0kg mass be suspended for the tension in the rope to be 177N? pivot point produced by this force? 4po 0.63 // 77(rsiA8 + E m1y T=lvg