I have been stuck on this problem for while thar area is 779.1 mm^2 sigma_x is 120.6497159 sigma_y is 0 and tau_xy is 0. Beta is 20 degrees -X Incorrect. Use the shear stress transformation equation, Equation Tnt = - (x − σy) sin@cose + Txy(cos²0 — sin²0)Tnt = -(ox - σy)sincose + Txy (cos²0 - sin²0). The angle 00 shown in the figure above is equal in magnitude tothe angle ßß given in the problem statement. Note that 00 turns clockwise from the xx axis, so when used in EquationTnt = - (ox - oy) sin cose + Txy (cos²0 – sin²0) Tnt = − (0₁-0₂)sin@cose + Txy (cos²0 - sin²0), 00 isnegative.Determine the shear stress parallel to the weld seams. Use the correct sign based on the sign convention for shear stress andconsistent with the n - t coordinate system shown.Tnt =i -38.776eTextbook and MediaHintBSave for LaterMPaPAssistance UsedUse the shear stress transformation equation, Equation T = -(ox-oy)sin@cose + Txy (cos²0 - sin²0).Attempts: 2 of 6 usedSubmit Answer

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Answered: n stuck on this problem for while thar…

n stuck on this problem for while thar area is 779.1 mm^2 sigma_x is 120.6497159 sigma_y is 0 and tau_xy is 0. Beta is 20 degree

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Engineering Drawing

Technical drawing has been around since ancient times. As in the ancient Renaissance, complicated drawings were created, including Leonardo da Vinci's drawing. Modern engineering drawing, including its specific orthographic projection with scale conventions, began in Europe (France) during the early stages of Industrialization. L.T. C. Rolt also made some of Marc's drawings for its block-making machinery (in 1799) that also may have contributed significantly to British engineering technology. Since he learned the art of presenting three-dimensional(3D) items in a two-dimensional(2D) plane, as we now term mechanical drawing. 

Question

I have been stuck on this problem for while thar area is 779.1 mm^2 sigma_x is 120.6497159 sigma_y is 0 and tau_xy is 0. Beta is 20 degrees

-
X Incorrect. Use the shear stress transformation equation, Equation Tnt = - (x − σy) sin@cose + Txy(cos²0 — sin²0)
Tnt = -(ox - σy)sincose + Txy (cos²0 - sin²0). The angle 00 shown in the figure above is equal in magnitude to
the angle ßß given in the problem statement. Note that 00 turns clockwise from the xx axis, so when used in Equation
Tnt = - (ox - oy) sin cose + Txy (cos²0 – sin²0) Tnt = − (0₁-0₂)sin@cose + Txy (cos²0 - sin²0), 00 is
negative.
Determine the shear stress parallel to the weld seams. Use the correct sign based on the sign convention for shear stress and
consistent with the n - t coordinate system shown.
Tnt =
i -38.776
eTextbook and Media
Hint
B
Save for Later
MPa
P
Assistance Used
Use the shear stress transformation equation, Equation T = -(ox-oy)sin@cose + Txy (cos²0 - sin²0).
Attempts: 2 of 6 used
Submit Answer

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why the expert here wrote -200 cos 75 when Σ Fx = 0 , I understand all steps but how he got - 200 cos 75 ( for the force ) (how the angle is 75 ) can you please clarify for me this point please ...... because I did not got the angle 75 dgree for 200 N force can you draw a diagrm for me please to see how 75 dgree links to 200 N