Already found: Angular acceleration of bar BC = 480rad.s-1 Moment = 15N.m CCW NE

Question

Already found:

Angular acceleration of bar BC = 480rad.s-1
Moment = 15N.m CCW

NEED TO FIND -

The components of the force exerted at B on bar BC.

Help??

The linkage ABCD is formed by connecting the 3-kg bar BC to the 1.5-kg bars AB and CD as shown. The motion of the linkage is controlled by the couple M applied to bar AB. At the instant shown bar AB has an angular velocity of 24 rad/s clockwise and no angular acceleration. Already found: Angular acceleration of bar BC = 480rad.s-1 Moment = 15N.m CCW NEED TO FIND - The components of the force exerted at B on bar BC.

Explanation / Answer

The bar BC applies force along BC at B
(the component of force perpendicular to bar AB) X (length of AB) = M
[as there is no angular acceleration]

fbc x {3002/(1252+3002)} x 0.125 = 15

fbc = 130 N

fbc perpendicular = 120 N

fbc along AB = 50 N

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