Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
9
Static Equilibrium; Elasticity and Fracture
Change chapter

9-1 and 9-2: Equilibrium
9-3: Muscles and Joints
9-4: Stability and Balance
9-5: Elasticity; Stress and Strain
9-6: Fracture
9-7: Arches and Domes

Problem 5
A
Fg=1.1×103NF_g=1.1 \times 10^3 N
Giancoli 6th Edition, Chapter 9, Problem 5 solution video poster
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COMMENTS
By Fatma on Sun, 1/1/2012 - 7:09 AM

ok

By Mr. Dychko on Fri, 8/24/2012 - 5:11 PM

Hi mattne, sorry for the late reply... summer holidays make things go a little slower...

FB isn't related to an angle since it's straight horizontal, so there's no need to separate it into components. It's all in the x-direction. I can see where you're coming from though, since there's an equation that says Fasin(45)=FbF_a \sin(45) = F_b. What this equation is saying is that the horizontal component of FA equals FB. The angle is related to FA, and it's used to find the horizontal component of FA. For equilibrium, all the horizontal components of all the forces have to be equal, but there's no trig. for FB since it's already horizontal.

Hope this helps, and good luck with your studies.

By mattne on Sat, 7/21/2012 - 4:39 PM

Why is sin used for FB? The picture just shows a straight line how do we relate the angle to FB?

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