Question

Cow force, Inclined plane Velocity, v B, slope DATA The coefficient of friction mu = 0.45 The initial velocity in the shown direction) is Vi = 50.8 ft/s stopping distance S = 27.4 feet USER SUPPLIED DATA FOR THE WEIGHT OF THE BOX = YOUR WEIGHT + YOUR HEIGHT IN INCHES Example - suppose your weight is 160 lbf and your height in inches is 170 inches, the weight W = 160 + 170 = 330 lbf FOR THE ANGLE OF SLOPE (IN DEGREES), ADD UP THE DIGITS OF YOUR ACADEMIC PROGRAM Example - suppose I am in Mechanical Engineering, the program code is 0504 - the sum would be 9 and the slope would be 8 degrees. CALCULATE Determine the required force to stop the cow in the allowed distance Determine the stopping time

Answer 1

271 lbf We have weight of cow, W = mg = 190 + 71 lbs Friction of coefficient, p= 0.45 slope angle, 0 = 5+3 = 8° gravity of acceleration, g = 32.2 ft/s?

FBD of of cow

oc bue jam bue Quisbuce e N + a

three forces acting, friction force (fr = uN), weight(mg) and F

a is the acceleration of cow

$Balancing\,\,forces\,\,in\,\,perpendicular\,\,to\,\,incline,N=mg\cos\theta$

$Balancing\,\,forces\,\along\,\,the\,\,incline,F-f_r-mg\sin\theta=ma$

    $F-\mu N-mg\sin\theta=ma\Rightarrow F=\mu mg\cos\theta+mg\sin\theta+ma.....................(1)$

$\Rightarrow F=0.45 (271)\cos8\degree+(271)\sin8\degree+\frac{271}{32.2}a$

$\Rightarrow F=158.5+8.4a...................................(1)$

$Initial\,\,velocity,V_i=-50.8\,ft/s$

$Stopping\,\,distance,s=-27.4\,ft$

$Final\,\,velocity,V_f=0$

$Using\,\,equation\,\,of\,\,motion\,\,V_f^2-V_i^2=2as$

$0^2-50.8^2=2a(-27.4)\Rightarrow a\approx47.092\,ft/s^2$

$Putting\,\,value\,\,of\, a\approx47.092\,ft/s^2\,\,in\,\,(1)$

  $\Rightarrow F=158.5+8.4(47.092)\Rightarrow \boldsymbol{F=554.3\,\,lbf}$