Acircular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°c and 100 kpa. determine the drag force acting on the sign. also determine the bending moment at the bottom of its pole whose height from the ground to the bottom of the sign is 1.5 m. disregard the drag on the pole. the drag coefficient for a thin circular disk (sign) is cd = 1.1..

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

= Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind =

L = Length of pole

Density

Drag force

Force on the circular sign is 147.7 N

Bending moment at the bottom of the pole is 221.55 Nm

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.