A force of $400 \mathrm{~N}$ is applied to the brake drum of $0.5 \mathrm{~m}$ diameter in a band-brake system as shown in the figure, where the wrapping angle is $180^{\circ} .$ If the coefficient of friction between the drum and the band is $0.25,$ the braking torque applied, in $N \cdot m$ is
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(A) 100.6
(B) 54.4
(C) 22.1
(D) 15.7
(A) 100.6
(B) 54.4
(C) 22.1
(D) 15.7
Question ID - 157482 | SaraNextGen Top Answer A force of $400 \mathrm{~N}$ is applied to the brake drum of $0.5 \mathrm{~m}$ diameter in a band-brake system as shown in the figure, where the wrapping angle is $180^{\circ} .$ If the coefficient of friction between the drum and the band is $0.25,$ the braking torque applied, in $N \cdot m$ is
(A) 100.6
(B) 54.4
(C) 22.1
(D) 15.7
A force of $400 \mathrm{~N}$ is applied to the brake drum of $0.5 \mathrm{~m}$ diameter in a band-brake system as shown in the figure, where the wrapping angle is $180^{\circ} .$ If the coefficient of friction between the drum and the band is $0.25,$ the braking torque applied, in $N \cdot m$ is
(A) 100.6
(B) 54.4
(C) 22.1
(D) 15.7
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