A 195.0-N sign is suspended from a horizontal strut of negligible weight. The force exerted on the strut by the wall is horizontal. Draw an FBD to show the forces acting on the strut. (Do this on paper. Your instructor may ask you to turn in this work.)The angle is 30 degree
Find the tension T in the diagonal cable supporting the strut.
N
I tried 195sin(30)= 97.5 N, but its wrong. Please someone help me out. Thank you. Sorry for re-posting.
3 answers
tension*sin30=195N
Because the force exerted on the strut by the wall is horizontal, you should try using 195.0(cos(30.0)) to get the tension since cos is adjacent over hypotenuse.
ΣFx: -Tsin30° + 200N = 0
The force (Tsin30) is acting in the opposite direction to the 200N. This has to be true in order for them to add to zero. If the forces are acting in opposite directions then they must have equal magnitudes for it to stay in place.
T = 400 N
The force (Tsin30) is acting in the opposite direction to the 200N. This has to be true in order for them to add to zero. If the forces are acting in opposite directions then they must have equal magnitudes for it to stay in place.
T = 400 N