Solucionario Beer And Jhonston

PROBLEM 8.1
Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when θ = 30o and P = 200 N.

SOLUTION
FBD block:

ΣFn = 0: N − (1000 N ) cos 30° − ( 200 N ) sin 30° = 0 N = 966.03 N

Assume equilibrium:
ΣFt = 0: F + ( 200 N ) cos 30° − (1000 N ) sin 30° = 0 F = 326.8 N = Feq.

But

Fmax = µ s N = ( 0.3) 966 N = 290 N Feq. >Fmax

impossible ⇒ Block moves

and

F = µk N
= ( 0.2 )( 966.03 N )

Block slides down

F = 193.2 N

PROBLEM 8.2
Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when θ = 35o and P = 400 N.

SOLUTION
FBD block:

ΣFn = 0: N − (1000 N ) cos35° − ( 400 N ) sin 35° = 0

N = 1048.6 N

Assume equilibrium:
ΣFt = 0: F− (1000 N ) sin 35° + ( 400 N ) cos 35° = 0

F = 246 N = Feq. Fmax = µ s N = ( 0.3)(1048.6 N ) = 314 N Feq. < Fmax

OK

equilibrium
∴ F = 246 N

PROBLEM 8.3
Determine whether the 20-lb block shown is in equilibrium, and find the magnitude and direction of the friction force when P = 8 lb and θ = 20°.

SOLUTION
FBD block:

ΣFn = 0: N − ( 20 lb ) cos 20° + ( 8 lb ) sin 20° = 0

N =16.0577 lb Fmax = µ s N = ( 0.3)(16.0577 lb ) = 4.817 lb

Assume equilibrium:
ΣFt = 0:

(8 lb ) cos 20° − ( 20 lb ) sin 20° − F
F = 0.6771 lb = Feq.

=0

Feq. < Fmax

OK

equilibrium
F = 0.677 lb

and

PROBLEM 8.4
Determine whether the 20-lb block shown is in equilibrium, and find the magnitude and direction of the friction force when P = 12.5 lb and θ = 15°.

SOLUTION
FBDblock:

ΣFn = 0: N − ( 20 lb ) cos 20° + (12.5 lb ) sin15° = 0

N = 15.559 lb Fmax = µ s N = ( 0.3)(15.559 lb ) = 4.668 lb

Assume equilibrium:
ΣFt = 0:

(12.5 lb ) cos15° − ( 20 lb ) sin 20° − F
F = 5.23 lb = Feq.

=0

but Feq. > Fmax impossible, so block slides up and
F = µk N = ( 0.25 )(15.559 lb )
F = 3.89 lb

PROBLEM 8.5
Knowing that θ = 25°, determine the range of valuesof P for which equilibrium is maintained.

SOLUTION
FBD block:

Block is in equilibrium:
ΣFn = 0: N − ( 20 lb ) cos 20° + P sin 25° = 0

N = 18.794 lb − P sin 25°
ΣFt = 0: F − ( 20 lb ) sin 20° + P cos 25° = 0

or Impending motion up: Therefore,
F = µs N ;

F = 6.840 lb − P cos 25°

Impending motion down: F = − µ s N

6.840 lb − P cos 25° = ± ( 0.3)(18.794 lb − P sin 25° )
Pup =12.08 lb Pdown = 1.542 lb

1.542 lb ≤ Peq. ≤ 12.08 lb

PROBLEM 8.6
Knowing that the coefficient of friction between the 60-lb block and the incline is µ s = 0.25, determine (a) the smallest value of P for which motion of the block up the incline is impending, (b) the corresponding value of β.

SOLUTION
FBD block (impending motion up)

φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.04°

(a)Note: For minimum P, P ⊥ R so β = φ s Then

P = W sin ( 30° + φ s )
= ( 60 lb ) sin 44.04° = 41.71 lb

Pmin = 41.7 lb
(b) Have β = φ s

β = 14.04°

PROBLEM 8.7
Considering only values of θ less than 90° , determine the smallest value of θ for which motion of the block to the right is impending when (a) m = 30 kg, (b) m = 40 kg.

SOLUTION
FBD block (impending motion to the right)

φs = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°

P W = sin φs sin (θ − φ s )

sin (θ − φs ) =

W sin φ s P
−1 

W = mg

(a)

m = 30 kg: θ − φ s = sin

 ( 30 kg ) 9.81 m/s 2   120 N

(

) sin14.036° 
 

= 36.499° ∴ θ = 36.499° + 14.036° or θ = 50.5°

(b)

 ( 40 kg ) 9.81 m/s 2  sin14.036°  m = 40 kg: θ − φs = sin −1    120 N  

(

)

= 52.474° ∴ θ =52.474° + 14.036° or θ = 66.5°

PROBLEM 8.8
Knowing that the coefficient of friction between the 30-lb block and the incline is µ s = 0.25 , determine (a) the smallest value of P required to maintain the block in equilibrium, (b) the corresponding value of β .

SOLUTION
FBD block (impending motion downward)

φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°

(a) Note: For minimum P, So and...