Ahungry 900 n bear walks out ona beam in an attempt to
retrieve some "goodies" hanging at the end(fig. p8.20). the beam is
uniform, weighs 200 n, and is 6.00 m long; the goodies weigh90.0
n.
when the bear is at x = 1 m, find the tension in
thewire and the components of the reaction force at thehinge.

(vertical reaction) (in n) = ?

The Tension in the cable is 392.3N. The vertical reaction at the hinge (fixed end) is 850.3N.

The tension in the cable was calculated using the principles of moments which states that the total sum of the clockwise moments is equal to the total sum of anticlockwise moments on a system in equilibrium.

Explanation:

Sum of Clockwise moments = Sum of anticlockwise moments

900×1 + 200×3 + 90×6 = T×Sin120 × 6

900 + 600 + 540 = 5.20T

T = 2040/5.20 = 392.3N

Summing vertical forces

R + Tsin120 = 900 +200+90

R + 392.3sin120 = 1190

R + 339.7 = 1190

R = 1190 - 339.7

R = 850.3N

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

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