alln Lab 10: Pre-Lab Questions 1. An equal arm balance is made by balancing a me

Question

alln Lab 10: Pre-Lab Questions 1. An equal arm balance is made by balancing a meter stick at its center, so it looks like a see-saw. The first rd mass, m1=3.0 kg, is hung at r, 0.25 m to the left ofthe center. Asecond mass, ma-1.4kg, is hung a r20.4 m to the right of center. At what location with respect to the center of the meter stick must a thi mass, m-0.8 kg, be hung for the meter stick to be in equilibrium? Hint: the angles are all 90°. (5 pts.) 2. A standard clothes hanger makes a pretty good equal arm balance. With nothing on it, t with the bottom side horizontal. (e) While hanging normally, where is the axis of rotation for this equal arm balance located? With respect to this axis, where is the center of mass located? (b) If you tip the hanger with your finger then release it, it swings back and forth until friction stops it a horizontal position. Use the concept of torque to explain why the hanger starts to swing w release it. Give your answers in a 3-5 well written sentences (5 pts.) 3. You hang a mass on the right side of an equal arm balance and then pull on the hanger with your hand to keep it balanced (a) If you pull on the left side of the balance do you have to pull up or down to keep the balance in equilib- rium? (b) If you pull on the right side of the balance do you have to pull up or down to keep the balance in equilibrium? (If you are unsure, try it with a clothes hanger!) Give your answers in a 3-5 well written sentences (5 pts.)

Explanation / Answer

since the system is in equilibrium ,then Net torque action on the system is zero

Torque due to m1 + torque due to m2 + torque due to m3 = 0

(m1*g*r1)-(m2*g*r2)+(m3*g*r3) = 0

(3*9.8*0.25)-(1.4*9.8*0.4)+(0.8*r3*9.8) = 0

r3 = -0.2375 m

right of the center at a distance r3 = 0.24 m me should be placed

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