1. UNIVERSIDAD NACIONAL DE SAN MARTIN
Curso: Mecánica de fluidos
Ejercicios de Hidrostática
Docente: Ing. Carlos Segundo Huamán
Torrejón
Alumno: Lleis Rojas Casique
Tarapoto – Perú
2021
2. 47)
Datos
ℎ = 50 𝑚
𝜌 = 1000
𝑘𝑔
𝑚3
𝑃 = 𝜌 ∙ ℎ ∙ 𝑔
𝑃 = (1000
𝑘𝑔
𝑚3 )(50 𝑚)(10
𝑚
𝑠2 )
𝑃 = 5 × 105 𝑁
𝑚2 ∙
1 𝐾𝑃
10𝑁
𝑃 = 5 × 104
𝐾𝑃 𝑚2
⁄
48)
Datos
ℎ = 30
𝐴 = 45 × 10−4
𝑚2
𝐹 =?
𝑃 = 𝜌 ∙ ℎ ∙ 𝑔
𝑃 = (1000
𝑘𝑔
𝑚3 )(30 𝑚)(10
𝑚
𝑠2 )
𝑃 = 3 × 105 𝑁
𝑚2 ∙
1 𝐾𝑃
10𝑁
𝑃 = 3 × 104
𝐾𝑃 𝑚2
⁄
𝑃 =
𝐹
𝐴
3 × 104
𝐾𝑃 𝑚2
⁄ =
𝐹
45×10−4𝑚2
𝐹 = (3 × 104
𝐾𝑃 𝑚2
⁄ )(45 × 10−4
𝑚2)
𝐹 = 135𝐾𝑃
3. 49)
Datos
𝐷𝑃. 𝑔 = 60 𝑐𝑚
𝑆. 𝑇𝑃. 𝑝 = 5𝑐𝑚2
𝐹𝑃 .𝑝 = 50𝐾𝑃
𝐹𝑃 .𝑔 =?
𝐻𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝐴
𝑃𝑝 =
𝐹
𝐴
𝑃𝑝 =
50𝐾𝑃
5𝑐𝑚2
𝑃𝑝 = 10𝐾𝑃 𝑐𝑚2
⁄
𝑡𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒
𝑃
𝑔 = 10 𝐾𝑃 𝑐𝑚2
⁄
𝑃
𝑔 =
𝐹
𝐴
10𝐾𝑃 𝑐𝑚2
⁄ =
𝐹
(30𝑐𝑚)2𝜋
𝐹 = (10𝐾𝑃 𝑐𝑚2
⁄ )(900𝑐𝑚2) ∙ (3.14)
𝐹 = 28 260𝐾𝑃
4. 50)
𝐴𝑐 = 𝜌𝑅 = 0.8 ⟶ 𝜌𝐴 = 800
𝑘𝑔
𝑚3
⁄
𝐴𝑙 = 𝜌𝑅 = 2.2 ⟶ 𝜌𝐴 = 2700
𝑘𝑔
𝑚3
⁄
𝑉 = (0.2𝑚)3
𝑉 = 8 × 10−3
𝑚3
𝑔 = 10𝑚
𝑠2
⁄
𝐸 = 𝜌𝐴𝑐 ∙ 𝑉 ∙ 𝑔
𝐸 = (800
𝑘𝑔
𝑚3 )(8 × 10−3
𝑚3 )(10
𝑘𝑔
𝑠2
⁄ )
𝐸 = 64𝑁 ∙
1𝐾𝑃
10𝑁
𝐸 = 6.4 𝐾𝑃
2700
𝑘𝑔
𝑚3
=
𝑚
8 × 10−3𝑚3
𝑚 = 21.6 𝑘𝑔
𝐹 = 𝑚𝑔 = (21.6𝑘𝑔)(10𝑚 𝑠2
⁄ )
𝐹 = 216𝑁 ∙
1 𝐾𝑃
10𝑁
𝐹 = 21.6 𝐾𝑃
𝑇 = 𝐹 − 𝐸
𝑇 = 21.6 − 6.4
𝑇 = 15.2 𝐾𝑃
𝐿𝑎 𝑙𝑒𝑐𝑡𝑢𝑟𝑎 𝑑𝑒 𝑙𝑎 𝑣𝑎𝑠𝑐𝑢𝑙𝑎 𝑒𝑠:
160 𝐾𝑃 + 6.4 𝐾𝑃
166.4 𝐾𝑃
mg
T
E
5. 51)
Datos
𝑉 = 85 × 10−3
𝑚3
21 𝐾𝑃 𝑦 7𝐾𝑃
70𝑁 = 𝜌𝐴𝑐 ∙ (85 × 10−3
𝑚3 )(10𝑚 𝑠2
⁄ )
𝜌𝐴𝑐 =
70
85 ×10−3(10)
𝜌𝐴𝑐 = 835 𝑘𝑔 𝑚3
⁄
𝜌𝑅 =
𝜌𝐴𝐶
𝜌𝐻2𝑂
⁄ =
835𝑘𝑔 𝑚3
⁄
1000 𝑘𝑔 𝑚3
⁄
𝜌𝑅 = 0.835
52)
Cuando cae al agua
𝜌𝑅 = 7,8 ⟶ 𝜌𝐴 = 7800
𝑘𝑔
𝑚3
𝑚𝑔 = 𝐸 − 𝑝𝑒𝑠𝑜
𝜌𝐻 . 𝑉. a = 𝜌𝐻2 𝑂.𝑉. 𝑔 − 𝜌𝐻 . 𝑉. 𝑔
a = (
𝜌𝐻2𝑂−𝜌𝐻
𝜌𝐻
)𝑔
a = (
1000
𝑘𝑔
𝑚3 −7800
𝑘𝑔
𝑚3
7800
𝑘𝑔
𝑚3
) 9.81
𝑚
𝑠2
a = −8,5
𝑚
𝑠2
Cuando cae al mercurio
a = (
𝜌𝐻2 𝑂 − 𝜌𝐻
𝜌𝐻
)𝑔
a = (
13500
𝑘𝑔
𝑚3 − 7800
𝑘𝑔
𝑚3
7800
𝑘𝑔
𝑚3
) 9.81
𝑚
𝑠2
a = 7,15
𝑚
𝑠2
6. 53)
Datos
𝐴𝑟𝑖𝑠𝑡𝑎 = 10𝑐𝑚 𝑃𝐴 = 7𝑘𝑝
𝑉 = 10−3
𝑚3
Cuando se sumerge en agua
𝐸 = (𝜌)(𝑉)(𝑔)
𝐸 = (1000
𝑘𝑔
𝑚3
) (10−3
𝑚3) (10
𝑚
𝑠2
)
𝐸 = 10 𝑁 .
1𝑘𝑝
10𝑁
= 1𝑘𝑝
𝑃𝑟𝑒𝑎𝑙 = 𝑃𝐴 + 𝐸 = 7𝑘𝑝 + 1𝑘𝑝
𝑃𝑟𝑒𝑎𝑙 = 8𝑘𝑝
Cuando se sumerge en glicerina
𝜌𝐴 = 1,26 ⟶ 𝜌𝑅 = 1260
𝑘𝑔
𝑚3
𝐸 = (1260
𝑘𝑔
𝑚3 ) (10−3
𝑚3 )(10
𝑚
𝑠2 )
𝐸 = 12,6 𝑁 .
1𝑘𝑝
10𝑁
= 1,26𝑘𝑝
El peso aparente
𝑃𝐴 = 𝑃𝑟𝑒𝑎𝑙 − 𝐸
𝑃𝐴 = 8𝑘𝑝 − 1,26𝑘𝑝
𝑃𝐴 = 6,74𝑘𝑝
54)
𝑃𝐸 = 𝜌𝑎𝑖𝑟𝑒. 𝑔 = 1,29
𝑘𝑝
𝑚3
𝜌𝑎𝑖𝑟𝑒 =
1,29
𝑘𝑝
𝑚3
10
𝑚
𝑠2
𝜌𝑎𝑖𝑟𝑒 = 0,129
𝑘𝑔
𝑚3
𝐸 = 𝜌𝑎𝑖𝑟𝑒. 𝑉. 𝑔
𝐸 = (0,129
𝑘𝑔
𝑚3 )(1000𝑚3 )(10
𝑚
𝑠2 )
𝐸 = 1290𝑘𝑝
𝐹𝑟 = 𝐸 − 𝑚. 𝑔 …… … . .1
𝑃𝐸 = 𝜌ℎ𝑒𝑙𝑖𝑜. 𝑔 = 0,18
𝑘𝑝
𝑚3
𝜌ℎ𝑒𝑙𝑖𝑜 =
0,18
𝑘𝑝
𝑚3
10
𝑚
𝑠2
𝜌ℎ𝑒𝑙𝑖𝑜 = 0,018
𝑘𝑔
𝑚3 ; 𝑚ℎ𝑒𝑙𝑖𝑜 = 𝜌ℎ𝑒𝑙𝑖𝑜. 𝑉
𝑚ℎ𝑒𝑙𝑖𝑜 = (0,018
𝑘𝑔
𝑚3 ) (1000𝑚3 ) = 18𝑘𝑔
𝐹 = (18𝑘𝑔)(10) = 180𝑘𝑝
En 1
𝐹𝑟 = 1290𝑘𝑝 − 180𝑘𝑝 = 1110𝑘𝑝
7. 55)
Datos
𝐴𝑖𝑟𝑒 = 0.50 𝐾𝑃
𝐴𝑔𝑢𝑎 = 0.30 𝐾𝑃
𝐵𝑒𝑛𝑐𝑒𝑛𝑜 = 0.32 𝐾𝑃
𝑃𝑎𝑟𝑎 𝑒𝑙 𝑎𝑔𝑢𝑎
𝑃𝑅 = 𝑃𝐴 + 𝐸
𝐸 = 𝑃𝑅 − 𝑃𝐴
𝐸 = 0.50 𝐾𝑃 − 0.30 𝐾𝑃
𝐸 = 0.20 𝐾𝑃 = 2𝑁
𝑃𝐴𝐿 = 0.50 𝐾𝑃 ∙
10𝑁
1 𝐾𝑃
𝑃𝐴𝐿 = 5𝑁
𝑚𝑔 = 5𝑁
𝑚(10) = 5𝑁
𝑚 = 0.5 𝑘𝑔
𝐸 = 𝜌𝑎𝑔𝑢𝑎 ∙ 𝑉𝑠 ∙ 𝑔
2𝑁 = (1000𝑘𝑔 𝑚3
⁄ )(𝑉𝑠)(10𝑚 𝑠2
⁄ )
𝑉𝑠 = 2 × 10−4
𝑚3
𝜌𝐴 =
0.5 𝑘𝑔
2 × 10−4𝑚3
= 2.500
𝑘𝑔
𝑚3
⁄
𝜌𝑅 =
𝜌𝐴
𝜌𝐻2𝑂
=
2500
1000
= 2.5
𝑃𝑎𝑟𝑎 𝑒𝑙 𝐵𝑒𝑛𝑐𝑒𝑛𝑜
𝐸 = 0.50 − 0.32 = 0.18 𝐾𝑃
𝐸 = 1.8𝑁
𝜌𝐵 ∙ (2 × 10−4)(10) = 1.8𝑁
𝜌𝐵 ∙ (2 × 10−3) = 1.8𝑁
𝜌𝐵 = 900𝑘𝑔 𝑚3
⁄
𝜌𝑅 =
𝜌𝐵
𝜌𝐻2𝑂
=
900 𝑘𝑔 𝑚3
⁄
1000𝑘𝑔 𝑚3
⁄
= 0.9
8. 56)
𝑒𝑛 𝑒𝑙 𝑎𝑖𝑟𝑒 = 3,372𝑝 = 3,372𝑥10−3
𝑘𝑝
𝑒𝑛 𝑒𝑙 𝑎𝑔𝑢𝑎 = 3,1468𝑝 = 3,1468𝑥10−3
𝑘𝑝
𝐸 = 3,372𝑥10−3
𝑘𝑝 − 3,1468𝑥10−3
𝑘𝑝
𝐸 = 0,4252𝑥10−3
𝑘𝑝 = 4𝑥10−4
𝑘𝑝𝑎𝑝𝑟𝑜𝑥
𝑚. 𝑔 = 3,372𝑥10−3
𝑘𝑝
𝑚 (10
𝑚
𝑠2 ) = 3,372𝑥10−3
𝑘𝑝 (
10𝑁
1𝑘𝑝
)
𝑚 = 3,372𝑥10−3
𝑘𝑔
Densidades
𝑏𝑟𝑜𝑛𝑐𝑒 ⟹ 𝜌𝑅1 = 8,8 ⟶ 𝜌𝐴1 = 8800
𝑘𝑔
𝑚3
𝑙𝑎𝑡𝑜𝑛 ⟹ 𝜌𝑅2 = 8,4 ⟶ 𝜌𝐴2 = 8400
𝑘𝑔
𝑚3
Hallando volúmenes
𝑏𝑟𝑜𝑛𝑐𝑒 ⟹ 𝑉1 =
𝑚
𝜌𝐴1
=
3,372𝑥10−3
𝑘𝑔
8800
𝑘𝑔
𝑚3
= 38𝑥10−8
𝑚3
𝑎𝑝𝑟𝑜𝑥
𝑙𝑎𝑡𝑜𝑛 ⟹ 𝑉2 =
𝑚
𝜌𝐴1
=
3,372𝑥10−3
𝑘𝑔
8400
𝑘𝑔
𝑚3
= 4𝑥10−7
𝑚3
𝑎𝑝𝑟𝑜𝑥
Hallando el Empuje Aproximado
𝑏𝑟𝑜𝑛𝑐𝑒:𝐸1 = (1000
𝑘𝑔
𝑚3 ) (38𝑥10−8
𝑚3)(10
𝑚
𝑠2 )
𝐸1 = 0,0038𝑁 = 38𝑥10−4
𝑁𝑎𝑝𝑟𝑜𝑥
𝑙𝑎𝑡𝑜𝑛: 𝐸1 = (1000
𝑘𝑔
𝑚3 )(38𝑥10−8
𝑚3)(10
𝑚
𝑠2 )
𝐸1 = 0,0004𝑁 = 4𝑥10−4
𝑁𝑎𝑝𝑟𝑜𝑥
𝐸𝑙 𝑟𝑒𝑠𝑜𝑟𝑡𝑒 𝑒𝑠 𝑑𝑒 𝑙𝑎𝑡𝑜𝑛
9. 57)
𝐸 = 𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜 𝑉𝑠𝑢𝑚𝑒𝑟𝑔𝑖𝑑𝑜 𝑔
sabemosque P=E
Entonces
𝑉𝑠𝑜𝑙𝑖𝑑𝑜 ∗ 𝐷𝑠𝑜𝑙𝑖𝑑𝑜 ∗ 𝑔 = 𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜 ∗ 𝑉𝑠𝑢𝑚𝑒𝑟𝑔𝑖𝑑𝑜 ∗ 𝑔
1(2650) = 13600(𝑥)
𝑥 = 0.19485
58)
Sabemosque KP=10N
g= 10m/𝑠2
sabemosque
𝑃ℎ20 = 𝑃𝑟𝑒𝑎𝑙 − 𝐸
60=100-E
E=40
Donde que
𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜𝐺𝑉 = 40
800(10)V=40
V=4/800
Ahora
P=mg
100=m(10)
m=10
donde
D=m/g
D=10/4/800
X
10. D=2000
Sabemosque ladensidadrelativaes
DR=2000/10000 =2
59)
𝑉
1 ∗ 𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜 = 𝑚𝑐 + 𝑚𝑏
𝑉2 ∗ 𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜 = 𝑚𝑐
(𝑉
1 − 𝑉2)𝐷𝑙𝑖𝑞𝑢𝑖𝑑𝑜 = 𝑚𝑏
entonces
𝑉
1 − 𝑉2 = 𝐴 ∗ 2 = 𝐿2 ∗ 2
Remplazandotenemos
𝐿2 ∗ 2 ∗ 100 = 2
𝐿 = √100
L=10
60)
𝑃
𝑐 = 0.5 ∗ 10−2 N
𝑃
𝑝 = 8.6 ∗ 10−2 N
𝑃𝑟𝑒𝑎𝑙 = 𝑃ℎ2𝑜 + 𝐸
0.5 ∗ 10−2 + 8.6 ∗ 10−2=7.1 ∗ 10−2 + 𝐸
E=2 ∗ 10−2
𝐷𝑙𝑉𝐺 = 2 ∗ 10−2
1000𝑉10 = 2 ∗ 10−2
𝑉 = 2 ∗ 10−6
Entoncestenemos
𝑃
𝑐 = 0.5 ∗ 10−2
𝑚𝑔 = 0.5 ∗ 10−2
𝑚 = 0.5 ∗ 10−3
𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝐷 =
𝑚
𝑣
D=250kg/𝑚3
12. Segunda parte del trabajo
CENTRO DE MASA DE UN CUERPO
El centrode masa se obtiene remplazandoW=mgy dW = gdm
13. 1. CENTROIDE
El centroide Cesun puntoel cual define el centrogeométricode unobjeto
El centroide coincideconel centrode masa o el centrode gravedadsolamentesi el
material eshomogéneo.
Si el objetotiene uneje de simetría,entoncesel centroidese encuentrafijoendicho
eje.
En algunoscasosel centroide nose encuentraubicadosobre el objeto.
