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A uniform slender rod of length $L=36$ in. and we…

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Problem 77 Easy Difficulty

A crate of mass $80 \mathrm{kg}$ is held in the position shown. Immediately after the man accidentally drops the right end of the crate, assuming that the box does not slip at point $E,$ determine ( $a$ ) the angular acceleration of the crate, $(b)$ the reaction at point $E$

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Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

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Video Transcript

Okay, so in this problem, they're asked, they give us this box that is on this ledge, Hang off this ledge and they ask us to compute some moments. So first of all, they want to know, Given this set up, what is the moment produced by the weight of the box? W about the point E. Uh, so the way we compute moment is forced times perpendicular distance. And so we know that the force of the way to the box is going to be MGI times the distance. And so they give us that the weight of the box is it 80 kilograms times 9.81 meters per second squared for gravity in a distance. So the distance is the tricky part. So we need the perpendicular distance from the axis of the force to the point E. So what we have here is when we're interested in this distance, here was green distance from that access to the point E. So we know that the bottom part of the box from DT is 0.85 meters and from B to the midpoint of the box 0.6 meters. So that gives us a distance for, uh, the green part of 0.85 minus 0.6, which is 0.25 meters in that if you plug, it'll into your calculator will give you a total moment of 196 point to Newton meters. Ah, and of course, with any moment when you know what the sense of the moment is. And so if we look ah, the direction of the force factor. We know this is gonna push our box in a clockwise direction so clockwise cool and then for a part B, they ask us, what is the minimum force applied at B. So we're looking at this point now to counter act the moment created by W. So in other words, we want a moment of equal magnitude and opposite sense. So the important thing to note here is that we need in orderto minimize the amount of force needed. We want to maximize perpendicular distance, right, because again, moment is forced times perpendicular distance. And so if we can maximize the distance, then we can minimize the force to the maximum distance. Here is going to be the high partners of this triangle. Ah, which you can get from the Pythagorean theorem, right? Is the square it? 0.5 squared plus zero point. Uh, sorry. This rate training plus 0.85 squared and we know that the moment needs to be 100 96.2. So the force, it's going to be the moment divided by the distance would used to say 1 96.2 Newton meters, divided by the square root of 0.5 squared plus 0 +85 squared, which is gonna be a quantity and meters. So the meat is gonna cancel. We'll get a quantity and Newton's, which is good. And if you play that on your calculator, you get 199 news. That's the magnitude of the force. Now we needed direction. So what we need is this forced to be applied a direction perpendicular to our access. And so in order for this to be perpendicular, we know that this angle we need the angle relative to the horizontal is going to actually be equal to this angle. Um, if you remember your trigonometry rules, right? And so to get this angle, we can just take Theo the Arc Tangent of 0.85 divided by 0.5. So we can say Saito, because Arc Tanne of Remember Tangent is opposite over adjacent. So are 10 of 28. 5.5 will give us this angle, and that is going to give us a value of 59.5 degrees, and that is that is going to be up from the horizontal, so that means that our direction is

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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

A uniform slender rod of length $L=36$ in. and we…

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Related Courses

Vector Mechanics for Engineers: Statics and Dynamics

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Video Transcript

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

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Farnaz Mohseni

Aspen Fenzl

Jared Enns

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

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Farnaz Mohseni

Aspen Fenzl

Jared Enns

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

Elyse Gonzalez

Farnaz Mohseni

Aspen Fenzl

Jared Enns

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

Vector Mechanics for Engineers: Statics and Dynamics