I\'m having a little trouble understanding how to find the moment of A. The cros

Question

I'm having a little trouble understanding how to find the moment of A. The cross product isn't making any sense to me. Any help would be appreciated. I am looking at the textbook Engineering Mechanics-Statics (6th) By Meriam. Chapter 3 problem 71. Specifically step 6 of the solution. I'm having a little trouble understanding how to find the moment of A. The cross product isn't making any sense to me. Any help would be appreciated. I am looking at the textbook Engineering Mechanics-Statics (6th) By Meriam. Chapter 3 problem 71. Specifically step 6 of the solution.

Explanation / Answer

In regards to step 6, here is the cross product explained: for (-1i-3j+1.5k) x (0i+0j-1962k) i= (-3*-1962)-(0*1.5)= -3*-1962= 5886 j=(0*1.5)-(-1*-1962)=-(-1*-1962)=-1962 k=(-1*0)-(-3*0)=0-0=0 for (-2i-6j+3k) x (Rcos(30)i+(P+Rsin(30))j+0k) noting that {Rcos(30) = sqrt(3)/2 = .866, and Rsin(30) = 1/2} i=(-6*0)-(3*(P+Rsin(30))=0-(3P+(3/2)R)=(-3P-1.5R) j=(Rcos(30)*3)-(-2*0)=3*(sqrt(3)/2R)= 2.59R or 2.6R k=(-2*(P+Rsin(30))-(-6*Rcos(30))= (-2P-(2*(1/2)R))+(6*(sqrt(3)/2)R)= (-2P-1R+5.196R) or (-2P-R+5.20R) Hope this helps, Brian

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