A uniform 8.00 m. 1500 kg beam is hinged to a wall and supported by a thin cable attached 2.00 m from the free end of the beam, as shown in the figure. The beam is fixed to the wall by a hinge at an angle of 30.0-above the horizontal. The angle between the cable and the beam is 40.0. 40.0 Starting from the force balance equation in the vertical and the torque balance equation around the point where the beam is fixed to the wall: What is the correct expression for the hinge force H? O (Mg-H sin 30) x 6m cos 30- Mg x 4m (Mg-H sin 30) x 6m - Mg x 4m OME Hsin 30) xóm Mexim (Mg-H cos 30) x 6m - Mg x 4m
A uniform 8.00 m. 1500 kg beam is hinged to a wall and supported by a thin cable attached 2.00 m from the free end of the beam, as shown in the figure. The beam is fixed to the wall by a hinge at an angle of 30.0-above the horizontal. The angle between the cable and the beam is 40.0. 40.0 Starting from the force balance equation in the vertical and the torque balance equation around the point where the beam is fixed to the wall: What is the correct expression for the hinge force H? O (Mg-H sin 30) x 6m cos 30- Mg x 4m (Mg-H sin 30) x 6m - Mg x 4m OME Hsin 30) xóm Mexim (Mg-H cos 30) x 6m - Mg x 4m
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter8: Rotational Motion
Section: Chapter Questions
Problem 90A
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Step 1: Vertical Force Balance Equation:
VIEWStep 2: Torque Balance Equation:
VIEWStep 3: Now, let's solve for T:
VIEWStep 4: Let's calculate the values of sin(40°) and cos(30°):
VIEWStep 5: Now, we can substitute the value of T back into the vertical force balance equation to find H:
VIEWStep 6: Finally, we can substitute the value of T we found earlier:
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