FR = 292 N
R = 126,8 °
R = 293,8°
B = 147,6°|
Günther Harm - Homepage |
| Ergebnislösungen zu Mt11.2.1 | |||
| 1 | KM: 1cm = 50 kN F1 = 275 kN F2 = 245 kN | ||
| 2 | a) |
FR = 292 N |
|
| b) |
|
||
| 3 | a) | FR = 30,98 kN R = 293,8° |
|
| b) | lR = 3,258 m | ||
| c) | FA = 20,527 kN FB = 14,75 kN B = 147,6° |
||
| 4 | FA = 508,25 N FB = 1390,2 N | ||
| Zurück zur Startseite | |||