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A bobbin of mass M = 3 kg has an inner cylinder of radius r = 6 cm and outer discs of radius R = 7 cm. It rests on a slotted incline (angle 37 deg) where friction prevents sliding. A block of mass m is connected via a cord wound on the inner cylinder and passing through the slot under the incline. If the bobbin is in static equilibrium, find m (in kg). (sin 37 deg = 0.6, cos 37 deg = 0.8, g = 10 m/s²)
- 21
- 7
- 9
- 11
Correct answer: 7
Solution
The cord is vertical (tension T = mg). Torque about axle centre C: T*r = f*R (friction at contact point). So friction f = T*r/R = m*g*6/7. Force balance along the incline (up positive): f + T*sin(37 deg) - Mg*sin(37 deg) = 0... correction: T acts into the slot (vertically down on block, upward on bobbin through cord), component along incline (upward) = T*cos(37 deg) is incorrect. Resolving the vertical tension T along and perpendicular to incline: component along incline (down the slope) = T*cos(90 deg-37 deg) = T*sin(37 deg)... let me redo. Cord exits slot vertically; incline at 37 deg. Component of T along incline = T*sin(37 deg) pulling down the slope (cord pulls cord-end down = pulls bobbin toward slot which is downhill). Component perpendicular = T*cos(37 deg) pushing into incline. Along incline: f (up) - Mg*sin(37 deg) (down) - T*sin(37 deg) (down)... this doesn't balance for m=7. The correct approach: torques about contact point. Torque of Mg about P = Mg*R*sin(37 deg). Torque of T (vertical, upward on bobbin inner cylinder) about P = T*(R*sin(37 deg)+r). Equilibrium: Mg*R*sin(37 deg) = T*(R*sin(37 deg)+r). Wait that gives m < M. Let me use the verified torque about C + force balance. f = Tr/R, along incline: f - Mg*sin(37 deg) + T*component_up = 0. If T is vertical-upward, its component up the incline = T*sin(37 deg)... then T*r/R - Mg*sin(37 deg) + T*sin(37 deg) = 0 => T(r/R + sin 37 deg) = Mg*sin(37 deg), T = Mg*sin(37)*R/(r + R*sin(37))... Checking m=7: T=70N, Mg*sin37=18N, T*(r/R+sin37) = 70*(6/7+0.6)=70*1.457=102 ≠ 18. That's wrong. Correct verified approach shows f = 60N, along incline: f - Mg*sin37 - T*cos53 = f - 18 - T*0.6 = 60-18-42=0. ✓ with T=70N, m=7.
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