Find the force required to barely lift one end of the 5.0m long. 100 kg log. Not
ID: 1973255 • Letter: F
Question
Find the force required to barely lift one end of the 5.0m long. 100 kg log. Note that when we lift one end of the log the normal force from the ground acts at the other end which is still on the ground. Note at well that 'barely lift' meant the object is still in rotational equilibrium. (490N) Find the force required to do a sit up. Assume the upper body has a weight of 400. N. The 1.20 cm is the muscle insertion and represents the distance between where the muscle is attached to the bone and where the bones pivot. (18 300 N) Find the net torque on the bicycle crank below if the rider pushes on the pedals with 350. N of force and there it 120. N of tension in the upper chain. Distances are shown on a second diagram for clarity. (86.6 Nm CW) A pendulum bob it suspended from the celling and pulled with a horizontal force of 2 N so that its cord makes an angle of 30 with the vertical. What is the tension in the pendulum cord? (4 N)Explanation / Answer
Question (6) Force applied at pedal F1 = 350 N The distance of this force from crank center, r1 = 27.5 cm The tension in the chain T = 120 N Distance of this force from crank center, r2 = 8.0 cm Then the torque acting on crank due to force applied at pedal = Frsin Since force and distance perpendicular to each other so sin = 1 Then 1 = F1 r1 = (350 N) (0.275 m ) = 96.25 Nm Now torque due to tension 2 = T r2 = (120 N ) (0.08 m) = 9.6 Nm Since one of the torque is in clockwise direction and other torque at counterclokwise direction. Therefore the net torque is = 1 - 2 = 96.25 Nm - 9.6 Nm = 86.65 NmRelated Questions
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