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The 3 -lb uniform rod $B D$ is connected to crank…

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Problem 124 Easy Difficulty

The 4 -kg uniform rod $A B D$ is attached to the crank $B C$ and is fitted with a small wheel that can roll without friction along a vertical slot. Knowing that at the instant shown crank $B C$ rotates with an angular velocity of 6 rad/s clockwise and an angular acceleration of $15 \mathrm{rad} / \mathrm{s}^{2}$ counterclockwise, determine the reaction at $A$

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Video Transcript

hello friends as cylinder figure AIVD brought off months for Katie on off Lent 400 millimeters that is quite for meter. Connect to the crank BC fitted with the small well that can roll without friction along a vertical slot. Given that at the instant so frank busy rotate with angular velocity off six radian per second. So angular velocity off crank VCs given six radiant per second in club by direction and angular acceleration is 15 radian per second square in counterclockwise direction. You have to calculate reaction at a Let us see it dynamics off, trying to busy. We point having got in the initial acceleration on a normal acceleration and with the acceleration Al Alfa and Angular Velocity, Omega Letter B C is given quite 1 m, so 10 initial acceleration off. The point is, we seem to help for quite one and two 15. That is 1.5 meter per second square and downward direction, not in intellect. Normal acceleration. Okay, off the point. Omega is square busy. Omega is given six radian per second squared. We see it find one, so it would be 3.6 meter per second squared towards left now. Econometrics up, Rod A TV. A punch acceleration at eight. This is angular acceleration of the road about center of gravity. They were into works 3.6 meters per second squared. This angle beater will be it. Angle will be sign in. Worse. Off we see upon a B. This is the point. The 60 point. Sorry, this is D point. And here is she Point V c is given quite one on a V is fine too. So theater will be 30 degree exploration off a in other direction. Acceleration off the plus acceleration off with respect to be exploration off B is 1.5 Don't work 3.6 towards right on With respect toe it will be point toe and for at angle theater if we take X component X component off zero So 3.6 minus trying to help course of 30. So from here Alfa, you will get 20.78 a radiant per second squared on direction is blocked by direction. No, let us hear Free body diagram off road This is the reaction at eight here the weight will act the X be right I m a X um, anyway and I into Elba. Calculating momento was about G point A. Into this distance, we have to resolve along it. You will get 0.1732 0.2 m cause up 30. It's called two I into help unless no moment of inertia. Many square by 20 in tow. Alfa, subjecting the value mass is given for K G lent his point. Food Al Pies 20.78 upon well in tow, one point 72 The reaction at a You will get 6.4 new turn towards left. That's all for it. Thanks for watching it.

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Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

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Problem

The 3 -lb uniform rod $B D$ is connected to crank…

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Related Courses

Vector Mechanics for Engineers: Statics and Dynamics

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Watch More Solved Questions in Chapter 16

Video Transcript

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

Liev Birman

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Vector Mechanics for Engineers: Statics and Dynamics

Liev Birman

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek Vector Mechanics for Engineers: Statics and Dynamics

Liev Birman

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics