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The block brake shown below is set by a spring that produces a force S on each arm equal to 682 lb. The wheel diameter is 14 in., and the angle of contact for each block is 110°. Using 0.40 as a value of the coefficient of friction between the drum and lining, determine the maximum torque in foot-pounds that the brake is capable of absorbing. S + S 1/2" 1/2"

The block brake shown below is set by a spring that produces a force S on each arm equal to 682 lb. The wheel diameter is 14 in., and the angle of contact for each block is 110°. Using 0.40 as a value of the coefficient of friction between the drum and lining, determine the maximum torque in foot-pounds that the brake is capable of absorbing. S + S 1/2" 1/2"

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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The block brake shown below is set by a spring that produces a force S on each arm equal to 682 lb. The wheel diameter is 14 in., and the angle of contact for each block is 110°. Using 0.40 as a value of the coefficient of friction between the drum and lining, determine the maximum torque in foot-pounds that the brake is capable of absorbing. S + S 1/2" 1/2"
Transcribed Image Text:The block brake shown below is set by a spring that produces a force S on each arm equal to 682 lb. The wheel diameter is 14 in., and the angle of contact for each block is 110°. Using 0.40 as a value of the coefficient of friction between the drum and lining, determine the maximum torque in foot-pounds that the brake is capable of absorbing. S + S 1/2" 1/2"
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can you please explain more and draw about summation of forces on this problem

= req Considering EM ↑ - = S(a+b) s(a+b) s(a+b) O 12 static equilibrium of the lever, sx (a+b) + fxr N = S = s (a + b) 2 (1-Meg) aximum torque, T = fr Nr-fr. Nr - Peq Nr (1-Peq) = Nr Nr Peq NXr 682 x (9½ +81
Transcribed Image Text:= req Considering EM ↑ - = S(a+b) s(a+b) s(a+b) O 12 static equilibrium of the lever, sx (a+b) + fxr N = S = s (a + b) 2 (1-Meg) aximum torque, T = fr Nr-fr. Nr - Peq Nr (1-Peq) = Nr Nr Peq NXr 682 x (9½ +81
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