Problem 523 A force of 400 lb is applied to the pulley shown in Fig. P-523. The pulley is prevented from rotating by a force P applied to the end of the brake lever. If the coefficient of friction at the brake surface is 0.20, determine the value of P.
Solution 523
$20f = 10(400)$
$f = 200 \, \text{ lb}$
$f = \mu N$
$200 = 0.20N$
$N = 1000 \, \text{ lb}$
$\Sigma M_A = 0$
$48P + 8f = 16N$
$48P + 8(200) = 16(1000)$
$48P = 14400$
$P = 300 \, \text{ lb}$ answer
Follow @iMATHalino
MATHalino