Solve Prob. 16.127 for θ=90^∘

Name: Problem 128, Chapter 16: Vector Mechanics for Engineers: Statics and Dynamics
Uploaded: 2020-09-28T13:04:57-08:00
Duration: 19 min 15 s
Channel: Khoobchandra Agrawal
Description: Solve Prob. 16.127 for θ=90^∘

Key Concepts

Trigonometric Identities

Trigonometric functions, especially sine and cosine, have well?known values at special angles such as 90°. Recognizing that sin(90°) is 1 and cos(90°) is 0 allows one to simplify expressions and equations that involve these functions. This concept is fundamental in many areas of mathematics and physics where the evaluation of trigonometric functions at specific angles can reduce the complexity of problems.

Substitution Method

The substitution method involves replacing a variable or parameter in a general equation with a specific value. In this context, substituting ? = 90° into a formula or equation transforms it into a simpler case, thereby facilitating a direct evaluation or solution. This method is a common strategy in solving equations, analyzing boundary cases, and verifying results.

Algebraic Simplification

Once variables are substituted with their specific values, the resulting expression often contains terms that can be simplified algebraically. This might involve eliminating zero terms, combining like terms, or reducing the expression to a more manageable form. Mastery of algebraic simplification is crucial across mathematics and physics as it allows for clearer interpretation and solution of problems.

Transcript

00:01 Hello, thanks as shown in the figure, there is a test trick was developed to perform fatigue testing on fitness trampolines.

00:10 A motor drives the 200mm radius flyway.

00:17 The radius of the flywheel av is given 200mm that is 0 .2 meters which is pinned at its center point in countertlock by direction with constant angular velocity of the angular velocity of av is given 120 rpm or 45 radiance per second and counter clock by direction.

00:54 The flywheel is attached to the slider cd by 400 mct connecting dot vc.

01:02 The length of vc is given 400mm or 0 .4 meter having the marks 5 kg.

01:23 And the mass of the link cd is given to kg at the extent when theta is 90 degree as shown in the figure we have to calculate reaction at c let us start performing it here we will draw the geometrical figure this is a flywheel a v a v and c this is the geometrical center of vc or center of mass this is given 0 .4 meters and this is given 0 .2 meter free body diagram of cd first we will draw the geometry which has to be used position vector of b with respect to a with the 0 .2 j cap meter position factor of c with respect to b 0 .4 j cap meter moment of inertia of vc about its center of mass 1 by 12 mvc and square vc so it can be written as substituting the value mass of vc is given 5 kg length of vc is 0 .4 square so it can be written as 1 by 15 kg meter square now we will draw the free body diagram of cd rod.

04:28 Free body diagram of cd dot.

04:40 This is c point having the reaction in y direction and along x direction.

04:56 This is the geometrical center of mass.

05:07 It has a mass of cd into acceleration of cd allowed y -axis.

05:25 And here is the fourth effects.

05:28 If we find submission of force along y -axis, then it will be minus.

05:44 It's a weight will act here, minus 2g, minus 2y, 2a, this is equation 1, 3 body diagram of vc rod, cy, cx, vx, vx, vy, this is the metrical center, of vc if it is having acceleration mvc x1 and the rotation in lock by direction so it will be ibc into alpha vc submission of fx can be written as cx plus bx to be mass of rod vc into acceleration of center of mass in excesses direction.

07:51 So from here we can write px to b.

07:55 Math of bc is 5 kg acceleration of bc minus cx say equation to summation of f y you will get c y plus b by minus 5g to be is equal to 5 into acceleration of bc along y x xx equation number 3 taking moment of force about g that must be equal to minus point 2 cx plus 0 .2 vx is called to i vc into alpha so it can be written as minus 0 .2 into x x x 0 .2 vx is called to alpha vc upon say equation number four.

09:21 Now velocity of b points we can measure.

09:30 Velocity of a, well as angular velocity of a v.

09:38 Cross position the cross vector of position vector of b with respect to a velocity of a zero plus omega a b k cap cross point to j cap...

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