Question

please do as many parts as possible. Thank you.

0.8 ft CM 3 1727 3.5 ft 13° 2. H0.42 43 A 258 pound safe containing $5 million dollars is on an incline. Derek is pushing on the safe with a force of P. EXAMPLE 6 ft History Part Description Answer Chk What is the minimum force P Try 3 - your answer of 107lb is not correct. required to prevent the box from sliding down the incline? (include units with answer) 4.27 pts.94% 107lb A. Format Check # tries : 3 Hide Details 2% try penalty Hints: 0,0 What is the minimum force P required to prevent the box from tipping downward? (include units with answer) 4.45 pts.98% 124lb B. # tries: 1 Show Details Format Check 2% try penalty Hints: 0,0 What is the minimum force P required to slide the box up the 4.55 pts.100 % C. # tries: 0 Show Details incline? Format Check (include units with answer) 2% try penalty Hints: 0,0 What is the minimum force P required to tip the box upward? (include units with answer) 4.55 pts 100 % D. # tries: 0 Show Details Format Check 2% try penalty Hints: 0,0

Answer 1

iver The mam m 258 pundo W-mg We know, W11.03 x9.8 1146.9 N ord A:The e p ouived do pevert down dhe indline i minimum P the box tom lsoling Psina P ConaCW coePsind Usine 3.SH 13 Pa i3o.42 46.9 Cal 43Psin3 wSine +fshd = lty6.9 sin 43 O.974 P +0.42 C838.79 .225 PJ O94 P 252.29 +o.095P-782.184 669 P-429,894 P: y02.15N Part B The minimum oce guived do povend Psin The moment gt A o P-- Paad C3.5)Ccme (0.42) Pood usine6) wsne 2.6ft Poo13 3.5) 3 Lo.4) sin 43 (2-6) 3.41P+o.30a= +3 w. 3.41P 47w 34!P> fyCliu6.a) NAhbh

Pant c The minium ce p uied to Ade He or u the indlhe in Psind Pood-sinet C04Psin Poad Pan 13=11469 Sin 3. + wSine wase Fs O.974 P 482.184 352:29 0 0945 P. 0 8795P = \134.474 P= 128991N Podt D oquired atee P The minimum the Jox upsard ia, Oc&A Psina Pasd x 3.5 = Psind xg 8 Pard 3.5f4 +W Sine x 2,6+ 2.6 wsine WCooe Co.42) PCO13 x3.5Psin 13 xo. 8 1146.9 Sin43 X 2. 6 146.9 Ca 3 (os 3.uIP 6.79 P +2033, 68 + 35229 3.231P=2385.97 P 438.46 N