A non uniform 80.0 g metalistic balances when the support is placed at the 51.0

Question

A non uniform 80.0 g metalistic balances when the support is placed at the 51.0 cm mark. At what location the metalistic should a 5.00-g tack be placed so that the stick will balance at the 50.0 cm mark? In the figure, a 10.0-m long bar is attached by africtionless hings to well and held horizontal by a pope that make an angle theta of 53 degree with the bar. the bar is uniform and weight 39.9 N. How far from the hinge should a 10.0 kg mass be suispended for the tenssion T in the rope to be 125 N?

Explanation / Answer

a) by balancing moment

80*1 = 5*x

x = 16 cm

b) by balancing forces

T*sin53 + Fy = 39.9 + 10*9.8

Fy = 37.9

by balancing moment

39.9*5 + 10*9.8*x = T*sin53*10

x = 8.15 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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