A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working. The student goes to th
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Answer:
B) a= 2.2 m/s², T₁ '= 5.45 N, T₂ ’= 3 N
8) F> T₁ ’> T₂’
Explanation:
A) In the adjoint we can see the free-body diagrams of each element and of the set
B)
i) The acceleration of the chain is
F - W = M a
M = m₁ + m₂ + m₃
a =
a = - g (1)
a = 9 / (3 0.250) - 9.8
a = 2.2 m / s²
the positive sign indicates that the system is rising
ii) the outside of the top link over the middle
T₁ '- W₂ -T₃ = m a
the acceleration of all the links is the same because they are united
Tension T3 is the lower link force that must be equal to its weight
T₃ = W₃
we substitute
T₁ ’= m a + W₂ + W₃
T₁ ’= m a + m g + m g
T₁ ’= m (a + 2g) (2)
T₁ ’= 0.250 (2.2 + 2 9.8)
T₁ '= 5.45 N
iii) the force on the lower link
T₂ ' -W₃ = m a
T₂ '= m a + m g
T₂ '= m (a + g) (3)
T₂ '= 0.25 (2.2 + 9.8)
T₂ ’= 3 N
1) The external forces are
set
* the force of the cueda (F) upwards
* the force of the drawer (W) down
Top link
* the strength of the rope (F)
* the pso of the link (W1)
* the tension of the second link (T2)
Middle link
* The tension of the first link (T1 ')
* Weight (W2)
* the tension of the last link (T3)
Lower link
* The tension of the intermediate link (T2 ')
* Weight (W3)
2) the action and reaction forces are
* T₂ and T₁ '
* T₃ and T₂ '
this are forces of equal magnitude and direction, applied in two bodies
3) in the attachment you have the free body diagram of the body, the vertical upward direction is considered positive
4) The inconnitas are the acceleration of the body and the internal tensions between each link bone 2 tensions
5, 6 and 7)
F - W₁ -T₂ = m a
T₁'- W₂ -T₃ = m a
T₂ '- W₃ = m a
T₂ = T₁ '
T₃ = T₂ '
One way to check that the sign is correct is that the action and reaction force between two bodies can cancel out when adding the equations
F -W₁ - W₂ - W₃ = (m + m + m) a
for which we can evaluate and find the acceleration of the systems
8) order the strengths from greatest to least
F = 9 N,
T₁ '= 5.45 N
T₂ ’= 3 N
F> T₁ ’> T₂’
9) repeat the problem for an exeran force of F = 7.35 N
the acceleration we substitute in equation 1
a = F / m -g
a = 7.35 / 0.75 - 9.8
a = 0
the system is in equilibrium
the voltages using 2 and 3 are
T₁ ’= m (a + 2g)
T₁ ’= 0.25 (0+ 2 9.8)
T₁ ’= 4.9 n
T₂ '= m (a + g)
T₂ '= 0.25 (0 +9.8)
T₂ ’= 2.45 N
the order of force is
F> T₁ ’> T₂’
Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
![T = \sqrt[4]{202.857 * 10 ^{8} }](https://tex.z-dn.net/?f=T%20%3D%20%20%5Csqrt%5B4%5D%7B202.857%20%2A%2010%20%5E%7B8%7D%20%7D%20)
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.