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Chrome File Edit View History Bookmarks People Window Help 9796 - 4) Sun 6:09 PM Lakshmi Seelam a E Assignments xMasteringPhysics: Week 9 x MasteringPhysics: Week 9x MasteringPhysics: Week 9x)Home | Chegg.com My Drive - Googie Drive x https://session.masteringphysics.com/myct/itemView?assignmentProblemID=53248626&offset-next; Apps Internship (New Yor College Physics 201 Week 9 (graded) Problem 10.62 Volunteer-Lifebeat D reading for 9/11 ear Internship Program Thieme Medical Pub Alzheimer's Disease . Project Healthcare / Other Bookmarks Help | Close Resources « previous 16 of 23 next» Problem 10.62 Submit My Answers Give Up A uniform 7.75 m, 1500 kg beam is hinged to a wall and supported by a thin cable attached 2.0 m from the free end of the beam, as shown in the figure below (Figure 1). The beam is supported at an angle of 30.0 above the horizontal Incorrect; Try Again; 3 attempts remaining Part B Find the tension in the cable Express your answer in newtons to two significant figures Submit My Answers Give Up Figure 1 of 1 Part C How hard does the beam push inward on the wall? Express your answer in newtons to two significant figures 40.0 2.0 m 30.0 Submit My Answers Give Up VI dback Continue

Explanation / Answer

q=10 deg, L= (7.75-2)= 5.75m

Applying Newton’s second law horizontally we get,

Fx- Tx =0 -------------(1)

Applying Newton’s second law horizontally we get,

Fy+Ty-mg =0 ---------------(2)

Applying law of conservation of torque at point of F,

T*L*sin40 – mg*L*sin30 =

T*5.75*sin40 – 1500*9.8sin30 = 0       => T= 3444.4 N

From (1)

Fx= Tx=Tcos10= 3444.4cos10 = 3392.0 N

From (2)

Fy = mg-Ty= mg-Tsin10 = 1500*9.8-3444.4sin10 = 14102 N

F= sqrt(Fx^2+Fy^2) = sqrt(3444.4^2+14102^2) = 14516.6 N

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