A former student of mechanics wishes to weight himself but has access only to a

Question

A former student of mechanics wishes to weight himself but has access only to a scale A with capacity limited to 100 lb and a small 20-lb spring dynamometer B. With the rig shown he discovers that when he exacts a pull on the rope so that B registers 19 lb, the scale A reads 67 lb. What is his correct weight? Ans. W - 162 lb The fixture AC is designed with screw adjustment to regulate the height of the 900-N load supported by the roller at A. The screw bears against a smooth surface at B. Calculate the magnitude of the total force supported by the pin at C. Neglect the weight of the fixture. Ans. C - 991 N

Explanation / Answer

here the weight of the man is divided into 3 parts (say W1, W2, W3) 1) one part acts on Scale A : As the reading on scale A is 67 lb, W1 = 67 lb 2) second for pulling the rope that hangs to B: As the reading on scale B is 19 lb, W2 = 19 lb 3) third for lifting the body weight partially using the pulley arrangement: we need to calculate this using the diagram given If we consider left portion of the complete arrangement, the scale B reads 19 lb So the tension in the string above is 19 lb this tension is carried through-out this string So now if we look at right side the four turns of the string represents a total force of 4T acting upward, Hence the total downward force at the right portion is 4T = 4 * 19 = 76 lb Hence the portion of weight that is lifted by pulley arrangement is W3 = 76 lb Now total weight of the man is W1 + W2 + W3 = 67 + 19 + 76 = 162 lb If you have any confusion or if you look for a clear diagram then please let me know i will send it to your inbox (profile).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.