Question

Mechanical Engineering Fundamentals 38MEF olut entif aw 3. A block weighing 120 kg is subjected to two forces as shown. Determine the unknown force P necessary to cause the block to accelerate at 0.5 m/s2 to the left. nee P= 548 N 130N 0.5m/s 130 N 10° P 120 kg To' (120x9 rin) -P u= 0.3 Fy = 0: NR + 130 SIN 10 - 120 x 9.81 = 0 ->Ff: 0.3RN NR -

Answer 1

Ans

FBD

0.5 m/s^2 130 N 10°7 120 kg AN 120*9.81 = 1177.2 N

$\sum F_y = 0$

→-1177.2 - (130 sin10) + N = 0

$\Rightarrow N =1199.77 N$

ΣF, = 0

$\Rightarrow (120*0.5) - P + (130cos10) + (\mu N) = 0$

$\Rightarrow (120*0.5) - P + (130cos10) + (0.3*1199.77) = 0$

$\Rightarrow P =547.96 N$ (ans)