How much tension must a cable withstand if it is used to accelerate a 1200-kg car vertically upward at ?
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This is Giancoli Answers with Mr. Dychko. This free body diagram shows the tension force pulling the car upwards, and gravity pulling the car downwards. And the net force is going to be the up force minus the down force. I always have my minus signs explicit in my formulas instead of having them putting plus signs here and then burying my negative within a variable. So, in case you're wondering why that's minus right there. So, this net force is going to equal ma, that's Newton's second law. And we can solve for the tension by adding Fg to both sides. And so, that's Ft equals ma + Fg. And Fg is mg. And there's a common factor m here which you can factor out just for fun. And we see that we have 1200 kilograms mass of the car times the 0.7 m/s squared acceleration that it has plus acceleration due to gravity, 9.8 m/s squared in this case 1.3x10^4 in Newtons is the tension force required from the cable.
Why is g positive? shouldn't it be negative, since the force it is pointing downward?
Hi sugarhuny20, this is a common question. In the end, the answer depends on personal preference and which convention you choose to follow. Typically is taken to be the magnitude of the acceleration of gravity, which means it is always a positive number. Both I, and the textbook, follow that convention. That means the acceleration due to gravity is if you have the typical coordinate system with down as the negative direction.
In the video for this solution the force of gravity was taken to be negative as you would expect, since it is directed down. That's the equation I'm referring to there. Notice the negative in front of the force due to gravity term. That negative takes care of the gravity's direction, so there is no need to introduce another negative when substituting for .
Hope that helps,
Mr. Dychko
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