ation is a 710 ft82 The bottle in the figure rides on a con

ation is a = 7.10 ft/82 The bottle in the figure rides on a conveyor belL Figure 1 lf 1hu bers acce hat the bottie is wide enough that twil sip on the belt without tipping Express your answer numerically to three significant figures. dete mine the minimum ooeficeni o static friction that prevents the bolle from sipping Assume Learning Goal: To apply the kinetic equations of motion to rigid bodes undergoing translation View Available Hints When a rigid bedy undergcee trarslation, each partic\'e of the body has the eame acceleration ag a, where ag is the acceleraticn of the center of mass, Also, the rotational equation of motion reduces toMG-0. The scalar equations of motion for rect lineer translation, where all patcles travel in paralel straight-ine paths, become 0.220 Correct wheru ? and ? are the sum of the foros in thu x ard y directions, respectively, m is lhe mass, and ? MG is the sum of the moments about the center of gravity. The scalar equations of mction for curvilinear translation, where all particles travel in parallel aurved paths, become Part B The door n the igure has height c = 3.00 m mass m = 10.9 kg and ter ofgravity e where b-150 m and d=1.50 m Figure 2 if the man exerts a force F-10.7 Nath=1.00 m off the ground, determne the amount of me. it takes the man to move the door a dstence 8-210 m to the right The door is initial y et est the initial elocity and poation are 8-0.000 m/s and 80 0.000 m, ruspectively. Exprees your answer numorically in soconds to throe significant figures. w Available Hint(s) where the subecrigts n and t denote the normal and tangential directons of motion, respectively. The momunt equation for both types of translation, ? = 0, can be replaced by a summation of moments abcu an arbitrary poin A, where the moment of mac must be accounted for with the following equation t= 2.07 where the ter )A is the moment of mag about point A. Corroct Part C Figure 1 of , te reaction forces at ports A and B Consider the samn door shown in Part B The distance of d-1.50 m is the dstance from the center of the door to the wheels at points A and ? Find A, and respectively Expross your answers numerically in newtons to three significant figures separated by a comma. w Available Hint(s) VEAvec A,. B Submit Incorrect; Try Again

Solution

Please upload figure (2) for the 1st question.
For 2nd question:
car mass(m)=1450kg
Rear wheel load=(m*b)/(a+b)=(1450*1.65)/(1.25+1.65) = 825kg
front wheel load =1450-825 =625 kg
Maximum horizontal thrust that can develop on the road = friction co-efficient*weight
for front wheel, it is =0.2*625*9.81=1225
acceleration = thrust / total mass =1225 / 1450 =0.845
so time = V / accelearation = (70*1000)/(3600*0.845) = 23 sec

for rear wheel it was , thrust =0.2* 825 *9.81 =1617
acceleration = 1617 / 1450 =1.115
time =(70*1000) / (3600*1.115) = 17.4 sec

please upload 1st figure