8. Calculate the total torque (magnitude and direction) for each of the followin

Question

8. 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.

(a) Here m = 5 g and d = 10 cm.

magnitude____________

direction [No direction, clockwise, or counter clockwise]

(b) Here m = 5 g and d = 10 cm.

magnitude_____________

direction [No direction, clockwise, or counter clockwise]

(c) Here m1 = m2 = 5 g and d1 = d2 = 10 cm.

magnitude_______________

direction [No direction, clockwise, or counter clockwise]

(d) Here m1 = m2 = 5 g and d1 = 10 cm and d2 = 0.9 m.

magnitude_______________

direciton [No direction, clockwise, or counter clockwise]

Explanation / Answer

Weknow that torque acting on a body si T = r X F = rF sin theta, r is lever arm or perpendicular distance from the point of application to the pivot point.
                               F is applied force ,theta is angle between force and lever arm

a) m = 5*10^-3 kg, d = 10 cm = 0.1 m

   T = 0.1*(5*10^-3*9.8)sin 90
= 0.0049 Nm in clock wise direction

b)
   m = 5*10^-3 kg, d = 10 cm = 0.1 m

   T = 0.1*(5*10^-3*9.8)sin 90
= 0.0049 Nm in anti clock wise direction

c)  
m1=m2 = 5g=5*0^-3 kg, d1=d2=10 c
T1 = 0.1*(5*10^-3*9.8)sin 90 =0.0049 Nm in anti clock wise direction, 0.0049 Nm in clock wise direction

   total torque is = zero

d)
m1=m2 = 5g=5*0^-3 kg, d1=10 cm=0.1 m,d2 =0.9 m

T1 = m1*g * d1*sin90 = 5*10^-3*9.8*0.1*1= 0.0049 Nm
T2 = m2*g*d2*sin theta = 5*10^-3*9.8*0.9*1= - 0.0441 Nm

net torque = 0.0049 - 0.0441= -0.0392 N m clockwise direction

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