Question

he desk has a weight of 80 lb and a centerof gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcomethe static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. Thecoefficients of static and kinetic friction at A andB are μs= 0.45 andμk= 0.25, respectively.

Answer 1

Answer:

$N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb$

Explanation:

Na and Nb are the vertical reactions on each of the two legs at A and at B

For the horizontal forces:

$Fcos(30)-0.5N_a-0.5N_b=0\\0.5N_a+0.5N_b= Fcos(30)\\N_a+N_b= 2Fcos(30)$

For the vertical forces:

$N_a+N_b-Fsin(30)-75=0\\N_a+N_b=Fsin(30)+80$

Therefore equating both equations:

$2Fcos(30)=Fsin(30)+80\\F(2cos(30)-sin(30))=80\\F=\frac{80}{2cos(30)-sin(30)} =64.93N$

After the desk star to slide:

sum of all vertical force = ma , therefore:

$N_a+N_b-64.93sin(30)-80=0\\N_a+N_b=64.93sin(30)+80$

sum of all horizontal force = ma

$64.93cos(30)-0.2N_a-0.2N_b=\frac{80lb}{32.2ft/s^2}a\\ 0.2(N_a+N_b)=64.93cos(30)-\frac{80lb}{32.2ft/s^2}a\\N_a+N_b=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}a}{0.2}=324.65-12.42a$

equating both equations:

$324.65-12.42a=64.93sin(30)+80\\12.42a=324.65-64.93sin(30)-80\\12.42a=212.185\\a=17.08ft/s^2$

From the moment equation:

$4N_b-80(2)-64.93(3)=\frac{-80}{32.2} (17.08)(2)\\N_b=67.48lb$

$N_a=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}(17.08)}{0.2}-67.48 = 45.04lb$

For each leg: $N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb$