A former student of mechanics wishes to weigh himself but has access only to a scale $A$ with a capacity limited to 110 lbs and a small 20 lb spring dynamometer $B$. With the rig shown, he discovers that when he exerts a pull on the rope so that $B$ registers 18 lbs, the scale $A$ reads 84 lbs. What is his correct weight?

Answer: Weight = _____ lbs

The correct weight of the former student is 102 lbs. The difference between the dynamometer reading and the scale reading is added to the scale reading to determine the correct weight.

To determine the correct weight of the former student, we can use the information provided about the readings on scale A and dynamometer B.

First, let's analyze the given information. When the student exerts a pull on the rope connected to dynamometer B, it registers 18 lbs. At the same time, scale A reads 84 lbs.

To find the correct weight, we need to understand the relationship between the readings on the dynamometer and the scale. The dynamometer measures the tension in the rope, while the scale measures the force exerted by the student's weight.

By comparing the readings, we can determine that the difference between the dynamometer reading and the scale reading is 18 lbs. This means that the weight of the student is greater than the tension in the rope by 18 lbs.

To find the correct weight, we can add the difference of 18 lbs to the scale reading of 84 lbs.

84 lbs (scale reading) + 18 lbs (difference) = 102 lbs

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

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