All calculations need the counting processand Free Body Diagram. 3. Determine the maximum force P that…

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Question “All calculations need the counting processand Free Body Diagram. 3. Determine the maximum force P that…”

All calculations need the counting processand Free Body Diagram. 3. Determine the maximum force P that... All calculations need the
counting processand Free Body Diagram.

3. Determine the maximum force P that can be applied without causing movement of the 300-lb crate that has a center of gravity at G. The coefficient of static friction at the floor is Is = 0.4. (20 Points) 1.5 ft 1.5 ft 2.5 ft P- ‘T 4.5 ft 3.5 ft

Answer

Giveno Free body diagram G 4.5 3.5′ W=300dbST NA W= 300lb NA = Normal force; FA = frictioned force… + Frichones Force 1″ to the normal forces.. FA = MgNA FA = 0.4NA

Efa=o+7 = 0.4×300=1 gold due to “FAFA=P Efy=0W = NA NA = 300lb P=FA=004NA 120lb But the location of “NA” will not exactly coincide with “W” because it should comenter bedance Klockwise instant 7 “NA acts right up to the “Gl Point P = -max n 11 5 N 9

W=300 lb A 2013 3.51 – FANA to EMG=0 (NA xe = (x) + (FA X 3-5), 300(e), (120x]]+(120×3.5), = 540 300e = 540 300 = 1.5 e=168ft right to 6 point

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