Question

The three forces acting on the beam can be replaced with a single equivalent force R. $P_1 = 25 ext{ kN}$ $P_2 = 60 ext{ kN}$ $P_3 = 80 ext{ kN}$ $40^circ$ $ heta$ $50^circ$ 4 m 6 m What is the magnitude of resultant force R? (kN) 133.45 None of the above -10.99 133.90 11.0

          The three forces acting on the beam can be replaced with a single equivalent force R.
$P_1 = 25 	ext{ kN}$ 
$P_2 = 60 	ext{ kN}$
$P_3 = 80 	ext{ kN}$
$40^circ$
$	heta$
$50^circ$
4 m
6 m
What is the magnitude of resultant force R? (kN)
133.45
None of the above
-10.99
133.90
11.0

The three forces acting on the beam can be replaced with a single equivalent force R.
P1 = 25 ext kN
P2 = 60 ext kN
P3 = 80 ext kN
40^circ
heta
50^circ
4 m
6 m
What is the magnitude of resultant force R? (kN)
133.45
None of the above
-10.99
133.90
11.0

Added by Juan Manuel B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Deepak Jangra

Step 1

The vertical component of P3 (P3y) can be found using the sine of the angle: P3y = P3 * sin(40) = 80 * sin(40) = 51.42 kN. The horizontal component of P3 (P3x) can be found using the cosine of the angle: P3x = P3 * cos(40) = 80 * cos(40) = 61.24 kN. Show more…

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The three forces acting on the beam can be replaced with a single equivalent force R. What is the magnitude of resultant force R? (kN) 133.45 None of the above -10.99 133.90 11.0