Calculate the total torque (magnitude and direction) for each of the following c

Question

Calculate the total torque (magnitude and direction) for each of the following cases shown below. You can assume the ''bar'' the masses are attached to is massless. The top point of the triangle represents the pivot point for the system. Be sure to include the magnitude and the direction. Here m = 6g and d = 18 cm. magnitude direction Here m = 6g and d = 18cm. magnitude direction Here m_1 = m_2 = 6 g and d_1 = d_2 = 18 cm. magnitude direction Here m_1 = m_2 = 6 g and d_1 = 18 cm and d_2 = 0.7 m. magnitude direction

Explanation / Answer

a)

F = weight = mg = 0.006 x 9.8

d = distance from axis of rotation = 18 cm = 0.18 m

Torque = mgd = 0.006 x 9.8 x 0.18 = 0.0106 Nm

b)

F = weight = mg = 0.006 x 9.8

d = distance from axis of rotation = 18 cm = 0.18 m

Torque = mgd = 0.006 x 9.8 x 0.18 = 0.0106 Nm

c)

F1 = weight = m1g = 0.006 x 9.8

d1 = distance from axis of rotation = 18 cm = 0.18 m

F2 = weight = m2g = 0.006 x 9.8

d2 = distance from axis of rotation = 18 cm = 0.18 m

Torque = m1gd1 - m2gd2 = 0.006 x 9.8 x 0.18 - 0.006 x 9.8 x 0.18 = 0 Nm

d)

F1 = weight = m1g = 0.006 x 9.8

d1 = distance from axis of rotation = 18 cm = 0.18 m

F2 = weight = m2g = 0.006 x 9.8

d2 = distance from axis of rotation = 0.7 m

Torque = m1gd1 - m2gd2 = 0.006 x 9.8 x 0.18 - 0.006 x 9.8 x 0.7 = - 0.031 Nm

magnitude = 0.031 Nm

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