Question

The 27.3-cm-diameter disk can rotate on an axle through its center. What is the net torque about the axle? (a=4.5 cm, q=41°, F1=25 N, F2=59 N, F3=30 N, F4=33 N.)

Answer 1

giventhe diameter of disk D = 27.3cmradis of disk = r = D/2 = 27.3cm/2 = 0.136mthe disk rotates on axle through its centerτ = F_Ar_Asinφ_A + F_Br_Bsinφ_B + F_Cr_Csinφ_C +F_Dr_D sinφ_D= F₃r_A sin ( -90) +F₁( 0.045m) sin90 +F₄ r_C sin135 +F₂r_D sin0= 30N( 0.136m) sin (-90) + 25N( 0.045) sin90 + 33N ( 0.045m) sin135 + 59N( 0.136m) sin0= -4.08 +1.125+1.0500 = - 1.90 N.ma negative torque means clock wise rotation of disk.

Answer 2

Radius of the disk r = D/2 = 27.3 / 2 = 0.1365 ma = 0.045 mTorque τ_1 = aF_1 sin270^0 = (0.045 m)(25 N ) (-1) = - 1.125 N.mTorque τ_2= rF_2sin(360- 41)^0 = (0.1365 m )( 59N)(-0.656) = - 5.283 N.mTorque τ_3 = rF_3 sin0^0 = 0.0 N.mTorque τ_3 = aF_4 sin(90 + 41)^0 = ( 0.045 m )(33 N )(0.7547) = 1.120 N.mnet torque τ = τ_1 + τ_2 + τ_3 + τ_4 = - 5.288N.mnegative sign indicates direction of rotation of disk which is clockwise