Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution
Check Mark

Answer to Problem 23P

  F=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

  F=0

Calculation:

  Physics for Scientists and Engineers, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

  F=Fhi^+fvj^(1)

Ignore the weight of the strut and use τ=0 at the hinge:

  LT(Lcos45)W=0

  T=Wcos45=(60N)cos45=42.43N

Use F=0 for the strut:

  Fx=FhTcos45=0

  Fy=Fv+Tcos45Mg=0

Solve for Fh :

  Th=Tcos45=(42.43N)cos45=30N

Solve for Fv :

  Fv=MgTcos45=60N(42.43N)cos45=30.0N

Subtitute in equation (1) to obtain:

  F=(30.0N)i^+(30.0N)j^

Conclusion:

  F=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution
Check Mark

Answer to Problem 23P

  F=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

  F=0τ=0

Calculation:

Including the weight of the strut, apply τ=0

  LT(Lcos45)W(L2cos45)w=0

  T=(cos45)W+(12cos45)w=(cos45)(60N)+(12cos45)(20N)=49.5N

Apply F=0 to the strut:

  Fx=FhTcos45=0

And

  Fy=Fv+Tcos45Ww=0

Solve for Fh :

  Th=Tcos45=(49.5N)cos45=35.0N

Solver for Fv :

  Fv=W+wTcos5=60N+20N-(49.5N)cos45=45.02N

Substitute in equation (1) to obtain:

  F=(35.0N)i^+(45.0N)j^

Conclusion:

  F=(35.0N)i^+(45.0N)j^

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