A crate of mass 80 kg is held in the postion shown. Determine ( a ) the moment p

Question

A crate of mass 80 kg is held in the postion shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E.

I know the correct answers: (a) 196.2 N*m i. (b) 321 N b 35.0 degrees. (c) 231 N x at point D.

I'll give you the pts if I can clearly follow your steps

A crate of mass 80 kg is held in the postion shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. know the correct answers: (a) 196.2 N*m i. (b) 321 N b 35.0 degrees. (c) 231 N x at point D.

Explanation / Answer

a) moment = force * distance in perpendicular direction = mg*(0.85-0.6) = 80g*0.25 = 196.2 N-m

b)let force be F in direction x degrees

moment due to F = moment atE = 196.2

Distance AE^2 = (1.2-0.85)^2 + 0.5^2 , so AE = 0.61

For F to be minimum distance should be maximum,

so distance should be 0.61

so F*0.61 = 196.2, so F = 321 N

for distance to be 0.61, foce should be applied in the direction normal to AE line,

so tan x = 0.35/0.5

so x = 34.99 = 35 degrees with horizontal

c)Let force be F

F*distance = moment at E = 196.2

so F*0.85 = 196.2

so F = 231 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.