Find the horizontal & vertical components: PH= 200 KN PV = 100 KN magnitude of P=√ (PH)² + (P₂) =(-200 kN)2+(100 kN) ² 40,000 + 10,00D 50,000 223.61 KN : Pv PH 0 = tan 100 KN -200 kN 0=- -26.57 b. FH = 300 KN Fy=-200 kN direction of P = tan magnitude of F = (FH)² + (FV) ² = 90,000 + 40,000 130,000 =360-56 kN (300 KN)²+(-200 kN) ² = -T Fy FH 200KN 300 kN direction of F = tan etan 10=-33.69⁰ Graphs:
Find the horizontal & vertical components: PH= 200 KN PV = 100 KN magnitude of P=√ (PH)² + (P₂) =(-200 kN)2+(100 kN) ² 40,000 + 10,00D 50,000 223.61 KN : Pv PH 0 = tan 100 KN -200 kN 0=- -26.57 b. FH = 300 KN Fy=-200 kN direction of P = tan magnitude of F = (FH)² + (FV) ² = 90,000 + 40,000 130,000 =360-56 kN (300 KN)²+(-200 kN) ² = -T Fy FH 200KN 300 kN direction of F = tan etan 10=-33.69⁰ Graphs:
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Draw the graphs for problems a and b provided in the picture.
Transcribed Image Text:Find the horizontal & vertical components:
a. PH: 200 KN
Py = 100 KN
b.
magnitude of P=√ (PH)² + (PU) ²
=√(-200 kN)²+(100 kN) ²
40,000 + 10, 000
direction of P = tan Pv
PH
100 kN
-200 kN
FH = 300 KN
Fv=-200 KN
50,000
223.61 KN
0 = tan!
0 = -26.57
magnitude of F
=√ (FH)² + (Fv 12
=√(300 KN)² + (-2₂00 kN) ²
90,000 + 40,000
=√130,000
=360-56 kN
direction of F = tan¹ Fv
FH
0 = tan 200KN
300 kN
0=-33.69⁰
Graphs:
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