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th4450
th4450 is offline
#12
Feb26-12, 07:42 AM
P: 38
Quote Quote by drawar View Post
F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
Quote Quote by drawar View Post
I am considering the two lower links (B and C) accelerating under the 3 forces.
If you are considering B and C, then it should not be 15 N there.