Any help with this two part question would be great! Any help with this two part

Question

Any help with this two part question would be great!

Any help with this two part question would be great! 005 (part 1 of 2) 10.0 points A 0.095 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0.657 kg object hangs vertically from the 5.75 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 18.3 N, find the value of the second mass. The acceleration due to gravity is 9.8m/s2. Answer in units of kg. 006 (part 2 of 2) 10.0 points Find the mark at which the second mass is attached. Answer in units of cm.

Explanation / Answer

The total mass to balance the string force is 18N / 9.8m/s^2 = 1.84 kg

So the second mass is 1.84 - 0.095 - 0.647 kg =1.098 kg

part 2
Equate clockwise and anticlockwise torques about the 40 cm mark.
Use distances relative to the 40 cm mark, and work in units of N.cm
Treat the mass of the meterstick as a point mass at the 50cm mark.

Unknown distance is d.
(Drawing a diagram helps!)

1.098*9.8*d + 0.095*9.8*(50-40) = 0.647*9.8*(40-5.38)

10.76d + 9.31 N.cm = 219.51 N.cm
d = 19.54 cm

Convert d into distance from 0 cm mark:

d = 40 + 19.54 cm =59.54 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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