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Forces in y-direction: [ R_y = W = 200 , N ]

So ( R = \frac200\sin\alpha = \frac200\sin 67.2° \approx \frac2000.922 \approx 216.9 , N).

So I = (2.5 cos50°, 5 sin50°).

Question: Trouvez les tensions ( T_1 ) et ( T_2 ) dans les câbles.

Forces in x-direction: [ R_x = T \quad (\textsince R \text has a horizontal component toward the right) ]

Also, moment equilibrium (or concurrency) gives: The line of ( R ) must pass through I.

Equilibre D 39-un Solide Soumis A 3 Forces Exercice Corrige Pdf Now

Forces in y-direction: [ R_y = W = 200 , N ]

So ( R = \frac200\sin\alpha = \frac200\sin 67.2° \approx \frac2000.922 \approx 216.9 , N). Forces in y-direction: [ R_y = W =

So I = (2.5 cos50°, 5 sin50°).

Question: Trouvez les tensions ( T_1 ) et ( T_2 ) dans les câbles. N). So I = (2.5 cos50°

Forces in x-direction: [ R_x = T \quad (\textsince R \text has a horizontal component toward the right) ] Forces in y-direction: [ R_y = W =

Also, moment equilibrium (or concurrency) gives: The line of ( R ) must pass through I.

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