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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Textbook Question
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Chapter 12.1, Problem 12.70P

Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the fixed circular slot DE of radius b = 20 in. Neglecting friction and assuming that θ ' = 15 rad/s and θ ' ' = 250 rad/s2 for the position θ = 20 ° , determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE.

  Chapter 12.1, Problem 12.70P, Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the

Expert Solution
Check Mark
To determine

(a)

The radial and transverse component of the resultant force at given condition.

Answer to Problem 12.70P

The required value of the radial component is 13.16lb and transverse component is 2.10lb.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.70P , additional homework tip  1

wB=4ozb=20inθ˙=15rad/sθ¨=250rad/s2θ=20o

Formula used:

ar=r¨rθ˙2

aθ=rθ¨+2r˙θ˙

F=ma

Calculation:

According to the geometry of the system we have.

r=2bcosθ.

r˙=(2bsinθ)θ˙.

r¨=2b(θ¨sinθ+θ˙2cosθ).

And, ar=r¨rθ˙2.

2b(θ¨sinθ+θ˙2cosθ)(2bcosθ)θ˙2.

2b(θ¨sinθ+2θ˙2cosθ).

2(2012ft)[(250rad/s2)sin20o+2(15rad/s)2cos20o].

1694.56ft/s2.

We know that, aθ=rθ¨+2r˙θ˙.

(2bcosθ)θ¨+2(2bθ˙sinθ)θ˙.

2b(θ¨cosθ2θ˙2sinθ).

2(2012ft)[(250rad/s2)cos20o2(15rad/s2)sin20o].

270.05ft/s2.

We know that, Fr=mar.

1/4lb32.2ft/s2×(1694.56ft/s2)

13.1565lb.

Fr=13.16lb

And also, Fθ=maθ.

1/4lb32.2ft/s2×(270.05ft/s2).

2.0967lb.

Fθ=2.10lb

Expert Solution
Check Mark
To determine

(b)

The forces P and Q exerted on pin B at given condition.

Answer to Problem 12.70P

The required value of the force is P=6.89lb and Q=14lb.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.70P , additional homework tip  2

wB=4ozb=20inθ˙=15rad/sθ¨=250rad/s2θ=20o

Formula used:

F=ma

Calculation:

The FBD is depicted below.

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.70P , additional homework tip  3

We know that, Fr=Fr.

Qcos20o.

Q=1cos20o(13.1565lb).

14.0009lb.

Now the sum of the angular forces.

Fθ=PQsin20o.

P=(2.0967+14.0009sin20o).

P=6.89lb.

Q=14lb

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