Question 10 of 10 < -/19 View Policies Current Attempt in Progress A horizontal, uniform beam (mass M and length L) is h

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Question 10 Of 10 19 View Policies Current Attempt In Progress A Horizontal Uniform Beam Mass M And Length L Is H 1 (80.74 KiB) Viewed 32 times

Question 10 Of 10 19 View Policies Current Attempt In Progress A Horizontal Uniform Beam Mass M And Length L Is H 2 (44.25 KiB) Viewed 32 times

Question 10 Of 10 19 View Policies Current Attempt In Progress A Horizontal Uniform Beam Mass M And Length L Is H 3 (37.7 KiB) Viewed 32 times

Question 10 Of 10 19 View Policies Current Attempt In Progress A Horizontal Uniform Beam Mass M And Length L Is H 4 (155.26 KiB) Viewed 32 times

Question 10 of 10 < -/19 View Policies Current Attempt in Progress A horizontal, uniform beam (mass M and length L) is hinged at its left end by a pin to a vertical wall. Its right end is connected to a massless cable that makes an angle 0 = 30° to the horizontal, as shown in the simulation (linked below). A uniform box (mass m) can be positioned anywhere along the length of the beam by the slider in the simulation, and the variable x denotes the position of the box as measured from the left end of the beam. The simulation shows five forces acting on the beam: the gravitational force on the beam itself (F.beam), the force from the box (F S.box), the tension from the cable (7), the horizontal force on the beam from the pin (PX). and the vertical force on the beam from the pin (,) Mouse over each blue arrow to see its label. Simulation Equilibrium (Beam and Cable) [1] This force is displayed to the side in the simulation, because it is not visible at the pin's actual location. Question 1 Position the box at the left end of the beam (x = 0). Use the plot of the tension magnitude to calculate the mass M of the beam. Neglect the size of the box. M- i kg Save for Later Attempts: 0 of 2 used Submit Answer

Question 10 of 10 < -/1 lii : Question 2 Adjust the position of the box such that x > 0. Use the displayed tension magnitude from the cable along with the mass of the beam M to find the mass of the box m. m = i kg Save for Later Attempts: 0 of 2 used Submit Answer Question 3 What is the magnitude of the horizontal force from the pin,Pa. when x = L? P N Save for Later Attempts: 0 of 2 used Submit Answer Question 4 What is magnitude of the vertical force from the pin, P., whenx = L?

Question 10 of 10 < -/1 E : Save for Later Attempts: 0 of 2 used Submit Answer Question 3 What is the magnitude of the horizontal force from the pin. Palu a. when x = L? N Save for Later Attempts: 0 of 2 used Submit Answer Question 4 What is magnitude of the vertical force from the pin, P., whenx = L? P = i N e Textbook and Media Save for Later Attempts: 0 of 2 used Submit Answer

x/L = 1.000 (N) 3500 320 2500 T = 3430, N 2000 UL 10 0.0 02 0.4 0.6 0.8 0 = 30° Px Equilibrium (Beam and Cable) A horizontal, uniform beam (mass M and length L) is hinged at its left end by a pin to a vertical wall. Its right end is connected to a massless cable that makes an angle 0 = 30° to the horizontal, as shown in this graphic. A uniform box (mass m) can be positioned anywhere along the length of the beam by the slider in the