For the arrangement at the night. Which tensional force(s) would create what is
ID: 1461494 • Letter: F
Question
For the arrangement at the night. Which tensional force(s) would create what is called counter clockwise torques? What tensional Cards) would creak what is called clockwise torques? If T1 = 50 N, what is the torque for this tensional force? ___ NM. If T2 = 40 N, what is the torque for this tensional force? ___ NM. What would T3 fuse to be for this system to be in equilibrium? T3 = N. For this figure, assume the meter stick is uniform and that the pivot is located at the 1 cm position. Is the torque created by the weight or the meter stick a clockwise or counter-clockwise torque? Is the torque created by the tension in the string a clockwise counter clockwise torque? If the string is attached to the meter stick at the 98 cm mark and is tied to the vertical rod 75 cm above the pivot, what would be the tension in the string for a mete rstick whoseExplanation / Answer
1. T3 will create counter clockwise torque because it is right to the axis and in up ward direction,. So the stick will go in counterclockwise direction because of T3.
2. T1 and T2 will cause clockwise torque.
3. T1=50 N, distance from axis, r= 0.4 m.
torque = r*T1 = 20 Nm
4. T2 = 40 N, r= 0.4m
torque = 40 N*0.4 m = 16 Nm
5. T1 = 50 N and torque due to T1 = 20Nm counterclockwise
T2 = 40 N and torque due to T2 = 16 Nm counterclockwise
therefore, torque due to T3 must be 20+16=36 Nm which already is in clockwise direction.
since, 36 Nm = T3*r3 = T3*(0.4+0.4)m
therefore, T3 = 36 Nm/0.8m = 45 N.
6. since weight is in downward direction right to the pivot. therefore, the torque due to weight of the meter stick is in clockwise direction.
7. string is keeping the stick in equilibrium. So, the string must apply a torque opposite to the weight of the stick. So the string is creating a torque in counter clockwise direction.
8. incomplete question.