Force and Free Body Diagram Practice Questions - The Burns Home ...

3.Figure 3y30N50F fF N5030cos 50x30sin 50FW60Na) From the free body diagram, since the box is moving with constant velocity, it is in equilibrium. Weknow that F 0, therefore:xTo two significant digits, F 19b) To find the normal force,NTo two significant digits, F 37c) From the definition of , we haveKf30Ncos 50F0NFff30Ncos 5019.28NF , we use the fact that Fy0fN30 sin 50 F N 60 N 0FNN37.02N Ff19.28NK0.52F 37.02NN4.Figure 4F T 470F T 570F T 3T1T2T 460 60520 5TT20T 33030F FT 1T 2500N30T 1T 230wTTBy symmetry 1 2 T T sin 60 T sin 60500N1 22 sin 60500N250NT sin 60288.675NThe tension in T 1and T 2is 289N