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Problem 13 in chapter 4 of the Giancoli Physics book askswhat the max and min fo

ID: 1740286 • Letter: P

Question

Problem 13 in chapter 4 of the Giancoli Physics book askswhat the max and min forces would be acting on a cable holdingan elevator (4850kg) at an acc. of .0680g.    Cramster's answer eventually sets up thisequation;       FTmin =m (-a +g)                  =4850(-.0680+1)g This +1 is placed in the equation for both the max andthe min and I really dont know where it came from or why.    Cramster's answer eventually sets up thisequation;       FTmin =m (-a +g)                  =4850(-.0680+1)g This +1 is placed in the equation for both the max andthe min and I really dont know where it came from or why. This +1 is placed in the equation for both the max andthe min and I really dont know where it came from or why.

Explanation / Answer

Ok, it took a minute but i think i figured it out. At first i looked at the acceleration as .0680 m/s^2. But, theacceleration isn't given directly in meters per second per second.It's a relative acceleration. (Relative to the acceleration due togravity) It's given as .0680 g...which means it's 680ten-thousandths of the value for g. (9.8 meters per second persecond). So, what they're doing in this step:     =4850(-.0680 +1)g is factoring out the gravitational constant g.4850(-.0680g+g)=4850(-.0680+1)g. It's just factoring. I hope that makes sense. If it doesn't, feel free to say soand i'll try to clarify it.