A former student of mechanics wishes to weigh himself but has access only to a scale A with capacity limited to 100 lbs and a small 24-lb spring dynamometer B. He discovers that when he exerts a pull on the rope so that B registers 22 lb, the scale A reads 80 lb. What is his correct weight?

Question

A former student of mechanics wishes to weigh himself but has access only to a scale A with capacity limited to 100 lbs and a small 24-lb spring dynamometer B. He discovers that when he exerts a pull on the rope so that B registers 22 lb, the scale A reads 80 lb. What is his correct weight?

Answer 1

To find the correct weight of the former student, we need to use the information given about the readings on scale A and spring dynamometer B.

Let's represent the unknown weight of the student as "W".

The scale A reads 80 lbs when the student exerts a pull of 22 lbs on the dynamometer B. This means that the weight W minus 22 lbs (the pull on dynamometer B) equals 80 lbs (the reading on scale A). So we can set up the following equation:

W - 22 = 80

Now, we can solve for W:

W = 80 + 22

W = 102

Therefore, the correct weight of the former student is 102 lbs.