Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 16.2, Problem 16.142P
To determine

Find a couple of M.

Expert Solution
Check Mark

Answer to Problem 16.142P

The couple of M=-7.19N.m.

Explanation of Solution

Given information:

The attached rod A mass is=0.8kg

And length is=160mm

Rod BP mass is=1kg

Length is=200mm

Explanation:

First consider the unit vector along the axis of X and Y direction.

So, we have to find the vertical distance from point P and point A(y)

y=[(200mm)1m1000mm]sin300=0.1m

Here, we take the coordinate points as (0,0.1m)

Then the position of the vector rP/A=(0.1m)j

Coordinate points of the point P with the point B is (0.2cos300m,0.2sin300m)

Then the position vector is rP/B=(0.1732m)i(0.1m)j

Then the coordinate points with the point P and point E as (0,-0.16m)

Position vector is rE/A=(0.16m)j

Here, the unit vector k as angular direction in clockwise is negative and it is positive when counter clockwise.

Angular velocity of the rod BP in vector form is

(20rad/s)k

Velocity of the rod BP is

vp=ωBP×rP/Bvp=[(20rad/s)k][(0.1732m)i(0.1m)j]=(3.464m/s)(k×i)+(2m/s)(k×j)=(3.464m/s)j+(2m/s)(i)=(3.464m/s)j(2m/s)i(1)

Formulate the velocity equation of point P

vp=ωAE×rP/A+vP/AEvp=(ωAEk)[(0.1m)j]+uj=(0.1ωAE)(k×j)+uj=(0.1ωAE)(i)+uj=(0.1ωAE)i+uj(2)

Compare the i terms from equation (1) And (2)

2m/s=0.1ωAEωAE=20.1=20rad/s

Compare the equation (1) and (2) for j terms. We get,

u=3.464m/s

So, the angular acceleration is zero.

Formulate the acceleration of the rod BP

ap=[(αBP)(rP/B)]ωBP2rP/Bap={(80rad/s2)k×(0.1732m)i(0.1m)j((20rad/s)k)2((0.1732)i(0.1m)j)}=(13.856)(k×i)+(8)(k×j)(400)((0.1732m)i(0.1m)j)=(13.856m/s2)j+(8m/s2)(i)+(69.32m/s2)i+(40m/s2)j=(13.856m/s2)j(8m/s2)(i)+(69.32m/s2)i+(40m/s2)j(3)

Find the acceleration of the point P to the point E

ap1=[(αAE)(rP/A)]ωAE2rP/Aap'={(αAEk)×(0.1m)j((20rad/s)k)2[(0.1m)j]}=(0.1αAE)(k×j)(400)[(0.1m)j]=(0.1αAE)(i)+(40m/s2)j=0.1αAEi+(40m/s2)j

Formulate the acceleration at the point P

ap=ap'+ap/AE+2ωBP×vp/AE

Substitute the above values

aP=0.1αAEi+(40)j+0+2[(20rad/s)k]((3.464m/s)j)=0.1αAEi+(40)j(138.56m/s2)(k×j)=0.1αAEi+(40)j(138.56)(i)=0.1αAEi+(40)j+(138.56)i(4)

Based on equations (3) and (4), we get,

(8m/s2)+(69.32m/s2)=0.1αAE+(138.56m/s2)αAE=77.2820.1αAE=772.82rad/s2

Find the weight of the rod AE is get by

WAE=mAE×gWAE=0.8kg×9.81m/s2=7.848N

Then the moment of inertial of the rod AE is

IAE=mAEIAE212IAE=0.8×(0.16m)212=1.7066×103kg.m2

Similarly, find the weight of the rod BP is

WAE=mAE×gWAE=1kg×9.81m/s2=9.81N

Then the moment of inertial of the rod BP

IBP=mBPIBP212IBP=1kg×(0.2m)212=3.333×103kg.m2

Here, the mass of the centers of rod AE and BP are to be considered as G and H.

Then the reference for point G with point A as (0,-0.08m).

Here the position vector is −(0.08m)j.

Similarly, the point H with point B as the reference as (-0.0866m,-0.05m). Position vector is −(0.0866m)i-(0.05m)j

Acceleration of the point G is

aG=αAErG/AωAE2rG/A

aG=[(772.82rad/s2)k][(0.08m)j][(20rad/s)k]2[(0.08m)j]=(61.8256m/s2)(k×j)+(32m/s2)j=(61.8256m/s2)(i)+(32m/s2)j=(61.8256m/s2)(i)+(32m/s2)j

Find the acceleration at point H

aH=αBPrH/BωBP2rH/BaH={((80rad/s2)k)[(0.0866m)i(0.05m)j][(20rad/s)k]2[(0.0866m)i(0.05m)j]}=(6.928)(k×i)+(4m/s2)(k×j)(400)((0.0866m)i)(0.05m)j=(6.928)(j)+(4)(i)(34.64)i+(20m/s2)j=(6.928)(j)(4m/s2)i(34.64m/s2)+(20m/s2)j=(30.64m/s2)i+(26.928m/s2)j

Vector Mechanics For Engineers, Chapter 16.2, Problem 16.142P , additional homework tip  1

Based on above figure, consider the moment about point B.

{((512)j)×P¯i+((0.3608)i(0.2083)j)×(2j)+Mk}={((0.3608)i(0.2083)j)×[(0.0621)((288.6)i+(166.64)j)+(3.5944×103)(0)]}

[0.4166P¯(j×i)+(0.7216)(i×j)+Mk]={((0.3608)i(0.2083)j)×[((17.930)i+(10.352)j)+0]}{0.4166P¯k+(0.7216)k+Mk}=3.7354k+3.735k(5) Substitute the values

{0.1(52.757)k+(0.8495)k+Mk}=0.8k0.2666k6.1252+M=1.0666M=7.19N.m

The magnitude of the couple applied on the rod is BP. It is considered as -7.19N.m and acts in a clockwise direction.

To determine

Find the force exerted on rod AE.

Expert Solution
Check Mark

Answer to Problem 16.142P

The force exerted by rod AE is F=52.757N.

Explanation of Solution

Vector Mechanics For Engineers, Chapter 16.2, Problem 16.142P , additional homework tip  2

Considering the above diagram of free body of the rod AE

rP/A×(P¯i)=IAEαAE+rG/A×(mAEaG)(0.1m)j×(P¯i)={(1.7066×103)((772.82)k)+((0.08m)j)×(0.8)[(61.8256)i+(32)j]}(0.1P¯)(j×i)=1.3188k+(3.9568)(j×i)(2.048)(j×j)0.1P¯k=1.3188k+3.8249(k)(2.048)(0)0.1P¯k=1.3188k3.9568k0.1P¯=1.31883.9568P¯=52.757N

The force exerted by rod AE by block P is 52.75N. It is acted upon the left direction.

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