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. With the rig shown 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?
thank you
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the weight of the former student is W.
According to the problem, when the student exerts a pull on the rope, the dynamometer B registers 22 lb and the scale A reads 80 lb.
We know that the dynamometer B weighs 24 lb, so when the student pulls on the rope, the additional force added to the system is 22 - 24 = -2 lb.
Now, let's set up the equations based on the given information:
Equation 1: W + B = A
Equation 2: W + (-2) = 80
Simplifying Equation 2, we have:
W - 2 = 80
Adding 2 to both sides of the equation, we get:
W = 82
Therefore, the correct weight of the former student is 82 lbs.
To find the correct weight of the former student, we can use the concept of equilibrium.
In this scenario, we have two forces acting on the system: the student's weight (unknown) and the tension in the rope. The tension in the rope is measured by the spring dynamometer B, which reads 22 lb.
When the student pulls the rope, the scale A reads 80 lb. This means that the total force acting on scale A is 80 lb, and we can break it down into two components: the student's weight and the tension in the rope.
Using the principle of equilibrium, we can set up the following equation:
Student's weight + Tension in rope = 80 lb
Since the tension in the rope is measured by dynamometer B and reads 22 lb, we have:
Student's weight + 22 lb = 80 lb
To find the student's weight, we can subtract 22 lb from both sides of the equation:
Student's weight = 80 lb - 22 lb
Simplifying, we get:
Student's weight = 58 lb
Therefore, the correct weight of the former student is 58 pounds.