1.A worker tries to move a rock by applying a 360-N force to a steel bar as shown. ( a ) Replace
that force with an equivalent force couple system at D. (b) Two workers attempt to move the
same rock by applying a vertical force at A and another force at D. Determine these two forces if
they are to be equivalent to the single force of part a . (Hibbeller)
Answers
Answer:
Explanation:
Omg
We can apply two forces at point D
Explanation:
R = 360N
θ = 50°
F = -360 Sin 40° i^ + (-360) Cos 40° j^
F = -231.4N i^ - 275.78 N j^
Equivalent moment:
YB/D = -(0.65 Cos 30° i^) + 0.65 Sin 30° j^
= -0.56 i^+ 0.325j^
Md = Yb/d → * F →
= (-0.56 icap + 0.325jcap) * (-231.4 icap - 275.78jcap)
= i^ j^ k ^
-0.56 0.325 0
-231.4 -275.78 0
= k^(-0.56 * -275.78) - (-231.4 * 0.325)
= 229.64 k^
Counter clockwise direction
360N = R
Angle from horizontal axis is 50.
Magnitude of the couple is 229.64.
B) R = 350Cos40 j^ + 360Sin49 i^
= F1 j^ + F2Cosθ (-i^) + F2Sinθ (-j^)
F2Cosθ = 360Sin40°
F2Sinθ = 360Cos40° - F1 = 23.24
By dividing 2 by 1, we get:
F2Sinθ / F2Cosθ = 23.24 / 360Sin40°
Tanθ = 23.24 / 360Sin40°
So θ = 5.735°
Substituting θ, we get: F2 = 231.4 / Cosθ =232.56 N
So we can apply two forces:
Vertical force of magnitude 252.54N
Another force at point D of magnitude 232.56N at an angle of 5.735 from horizontal.
