If a hook can sustain a maximum withdrawal force of 250 N inthe vertical direction, determine the maximum tension, T, that can be exerted.the vector T, points down in the fourth quadrant with the inside angle at 10 degrees.

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It isn't clear what is meant by "the inside angle is 10 deg" so we will solve this both ways.If the force is mainly sideways, only a small component is down and we can have a tension much larger than 250N. Sin(10) is very smallTsin(10) can be no more than 250NT=250N//sin(10)=1,440NThe other way to interpret "the ... See the full answer