Ok, so this is what I did: Left side of stick: Force = (9.79)(0.19)= 1.8601 N di

Question

Ok, so this is what I did:

Left side of stick: Force = (9.79)(0.19)= 1.8601 N      distance left: 60cm - 25 cm = 35cm = 0.35m

T left = F * d so T = 1.8601 N * 0.35m = 0.651 Nm     

Right side of stick: Force = (9.79)(0.37)= 3.6223N

T right= 3.6223N * 0.25m = 0.906 Nm

So the aparaturs would not be at equlibrium. But how do I get the net torque. Did I do this right?

Given the situation in the figure. The mass m1 is 0.58 kg and it is located at x1 = 25 cm. The pivot point is represented by the solid triangle located at x = 60 cm. The mass of the meter stick (m ms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.16 kg and it is located at x2 = 85 cm. Calculate the net torque (in N?m with the proper sign) due to these three weights. Use g = 9.79 m/s2. Ok, so this is what I did: Left side of stick: Force = (9.79)(0.19)= 1.8601 N distance left: 60cm - 25 cm = 35cm = 0.35m T left = F * d so T = 1.8601 N * 0.35m = 0.651 Nm Right side of stick: Force = (9.79)(0.37)= 3.6223N T right= 3.6223N * 0.25m = 0.906 Nm So the aparaturs would not be at equlibrium. But how do I get the net torque. Did I do this right?

Explanation / Answer

Net torque : = m1.g.(60-25).10-2 + mms.g.(60-50).10-2 - m2.g.(85-60).10-2

                 = 0.58.(9.79).(35x10-2) + 0.40.(9.79).(10x10-2) - 0.16(9.79).(25x10-2)

                 = 1.98737 N.m with direction counterclock wise

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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