Solve Prob. 3.17 assuming that the direction of 𝐓C is reversed.

Name: Problem 18, Chapter 3: Mechanics of Materials
Uploaded: 2020-09-07T12:21:01-08:00
Duration: 6 min 11 s
Channel: Khoobchandra Agrawal
Description: Solve Prob. 3.17 assuming that the direction of 𝐓C is reversed.

step 1 Hi everyone here it is given a disk of radius 180 mm. This is the radius of the disk which is 18

Solve Prob. 3.17 assuming that the direction of 𝐓C is...

Key Concepts

Static Equilibrium

Static equilibrium involves ensuring that the sum of all forces and the sum of all moments acting on a system are zero. This concept is essential in determining how reversing a force such as a tension affects the overall balance of the system.

Sign Conventions and Force Direction

Using consistent sign conventions and correctly assigning the direction of forces is fundamental in mechanics. Reversing the direction of a force, like the tension force in this problem, requires careful adjustment of the signs in the equations, which in turn impacts the calculated magnitudes and directions of other forces.

Free Body Diagram

A free body diagram is a graphical representation of a mechanical system isolated from its surroundings, showing all the forces acting on it. In problems where force directions change, such as with a reversed tension force, accurately drawing and labeling the free body diagram is crucial for setting up the equilibrium equations correctly.

System of Equations in Mechanics

Once the free body diagram and the correct sign conventions are established, the balance of forces and moments leads to a set of simultaneous equations. Solving these equations provides the unknown forces or reactions. This process is at the heart of many mechanics problems, especially when modifications such as a reversed force are introduced.

Transcript

00:01 Hi everyone here it is given a disk of radius 180 mm.

00:07 This is the radius of the disk which is 180 millimeter that is 0 .18 meter is at rest when it is placed so initially it is at rest when placed in contact with the belt moving downward with speed be neglecting the weight of the link a -v to be negligible coefficient of friction between the belt and the disk is 0 .4 we have to find angular acceleration when silping starts this will be the direction of normal reaction there will be a friction and its weight m g now let us start performing see the free body diagram again friction on the disc will be in this direction on the well it will be in the upper direction normal reaction in this direction weight vertically downward on the disk so let us see here submission of forces along a horizontal direction must be zero because it is in translatory equilibrium so n minus f a v there will be a force i forget to write here the force on a will be in this direction along a v direction i have marked wrongly just a moment this will be the direction of force now and minus a v f a v cause of theta so from here you write f a v cause of theta to be n first equation now summation of force along by axis upward we are taking positive that must be also zero so f a v sign of theta minus m g minus friction and friction will be mu k into n so from here we can write f a b to be f a v sine of theta to be m g plus mu k and divide equation 2 by 1 so you will get 10 of theta to be m g plus mu k and divide equation 2 by 1 so you will get 10 of theta to be to be m g plus mu k n upon n so from here value of n you will get m g upon 10 of theta minus mu k so you may calculate the friction the friction will be mu k n that is mu k m g divided by 10 theta minus mu k so this is equation now due to the friction it will experience the disc experience the angular acceleration so moment of force on that is about the point a you will get friction into r is equal to i into alpha so from here you may write alpha to be r upon i into f r is this moment of the disk about its perpendicular axis mr square and forces mu k m g divided by 10 of theta minus mu k simplifying so you will get alpha to be 2g up on r into mu k upon 10 of theta minus mu k now substituting the value g is 9 .8 radius is 0 .18 coefficient of friction is 0 .4 10 of 60 minus 0 .4...

Please give Ace some feedback