5f56f3b9-4929-41c8-925d-e61e0978011e 7 4 5 decimal

&[DATE] &[TIME] - &[FILENAME]

Beispiel Schiefe Ebene (Böge B338)

Gegeben:

F.g 1 kN * :

α 7 ° * :

μ 0.16 :

μ.0 0.19 :

Gesucht:

Die waagrechten Kräfte F1 - F3 für folgende Fälle:

F1 (bei gleichförmiger Bewegung nach oben)

F2 (bei gleichförmiger Bewegung nach unten)

F3 (Minimalkraft damit das Objekt nicht nach Unten rutscht)

Version 1: Calculating by the use of the function solve

Gleichgewichtsbedinungen:

ΣFx=0=Fx-Fgx-Fr1 ΣFy=0=-Fy-Fgy+Fn1 Fr1=μ*Fn1 Fgx=Fg*sin(α) Fgy=Fg*cos(α) Fx=F*cos(α) Fy=F*sin(α)

eq 1 el F.1x F.gx - F.r1 - 0 ≡ (:

eq 2 el F.1y - F.gy - F.n1 + 0 ≡ (:

eq 3 el μ F.n1 * F.r1 ≡ (:

eq 4 el F.gx F.g α sin * ≡ (:

eq 5 el F.gy F.g α cos * ≡ (:

eq 6 el F.1x F.1 α cos * ≡ (:

eq 7 el F.1y F.1 α sin * ≡ (:

eq F.1 F.1x F.1y F.gx F.gy F.n1 F.r1 71 mat Solve F.1 F.g α cos μ * α sin + (* α sin μ * α cos - / - (≡ (F.1x α cos F.g * α cos μ * α sin + (* α sin μ * α cos - / - (≡ (F.1y α sin F.g * α cos μ * α sin + (* α sin μ * α cos - / - (≡ (F.gx α sin F.g * ≡ (F.gy α cos F.g * ≡ (F.n1 F.g α sin μ * α cos - / - (≡ (F.r1 μ F.g * α sin μ * α cos - / - (≡ (71 sys

eq Unknowns F.1 F.1x F.1y F.g F.gx F.gy F.n1 F.r1 α μ 101 mat

I try to calculate just two variables (Fgx and Fgy) (I know i could just solve them but i try to understand how to use the different methods to solve equations):

eq_fg 1 el Fgx : Fg α sin * ≡

eq_fg 2 el Fgy : Fg α cos * ≡

eq_fg Unknowns Fgx Fgy 21 mat

Fgx Fgy Clear 1

Gew eq_fg Fgx Fgy 21 mat solve : #

Fgx Fgy Clear 1

eq_fg Fgx 100 N * ≡ Fgy 100 N * ≡ 21 mat FindRoot 0021 mat

Why is it 0??

Fgx Fgy Clear 1

eq_fg 1 el Fgx 0 N * ≡ 1000 N * Bisection 5.9605 105 -^* N *

the result is not correct! if i choose another start value there will be no result at all

Fg #

Fgx Fg α sin * :

Fgx #

Version 2: Calculating step by step using SMath as a formel editor.

ΣFx=0=F1x-Fgx-Fr1 ...eq1 ΣFy=0=-F1y-Fgy+Fn1 ...eq2 Fr1=μ*Fn1 ...eq3 Fgx=Fg*sin(α) ...eq4 Fgy=Fg*cos(α) ...eq5 F1x=F1*cos(α) ...eq6 F1y=F1*sin(α) ...eq7

is there a way to substitute eq3 into eq1 automatically? or can I rearange eq1 to Fr1=Fx-Fgx automatically? I can do it step by step but for larger formulas its a lot of typing.

Fgx Fg α sin * :

Fgx #

Fgy Fg α cos * :

Fgy #

F1x F1 α cos * :

F1y F1 α sin * :

0 F1x Fgx - Fr1 - ≡

...eq1

Fr1 μ Fn1 * ≡

...eq3

e.g.:here I would like to substitute Fr=μ*Fn and then rearange automatically to Fn1=...... Is there a way to do this? I read that sym- bolical treatment is difficult.

eq 3 in eq 1....

0 Fx1 Fgx - μ Fn1 * - ≡

Fn1 Fx1 Fgx - μ / ≡

In 2 einsetzen:

0 Fx1 Fgx - μ / Fy1 - Fgy - ≡

Fgy Fgx μ / + F1 α cos * μ / F1 α sin * - (≡

Fgy Fgx μ / + F1 α cos μ / α sin - (* ≡

F1 Fgy Fgx μ / + α cos μ / α sin - (/ :

F1 #