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2 years ago

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An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a lengt

h of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)

1 answer:

ehidna [41]2 years ago

3 0

The the tension in the horizontal rope is 7,019.4 N.

<h3>What is tension?</h3>

Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.

Tension can also be described as the action-reaction pair of forces acting at each end of said elements.

The unit of tension is Newton (same unit as force, since tension is also a force).

<h3>Tension in the horizontal rope</h3>

The tension in the horizontal rope is calculated as follows;

Apply the principle of torque;

T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ

T = (M + 2m)g sinθ/cosθ

T = (M + 2m)g tanθ

<em>let θ = 66⁰ (this value should be given in the question)</em>

<em />

T = (M + 2m)g tanθ

T = (74.9 + 2 x 122)(9.8)(tan 66)

T = 7,019.4 N

Thus, the the tension in the horizontal rope is 7,019.4 N.

Learn more about tension here: brainly.com/question/918617

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3 years ago

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Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o

Harman [31]

Answer:

A) 1.55

B) 1.55

C) 12.92

D) 34.08

E) 57.82

Explanation:

The free body diagram attached, R is the radius of the wheel

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.

At the centre of the whee, torque due to B is given by

${\tau _2} = - {T_{\rm{B}}}R$

Similarly, torque due to A is given by

${\tau _1} = {T_{\rm{A}}}R$

The sum of torque at the pivot is given by

$\tau = {\tau _1} + {\tau _2}$

Replacing ${\tau _1}$ and ${\tau _2}$ by ${T_{\rm{A}}}R$ and $- {T_{\rm{B}}}R$ respectively yields

$\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}$

Substituting $I\alpha$ for $\tau$ in the equation $\tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

The angular acceleration of the wheel is given by $\alpha = \frac{a}{R}$

where a is the linear acceleration

Substituting $\frac{a}{R}$ for $\alpha$ into equation

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ we obtain

$\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

Net force on block A is

${F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}$

Net force on block B is

${F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g$

Where g is acceleration due to gravity

Substituting ${m_{\rm{B}}}a$ and ${m_{\rm{A}}}a$ for ${F_{\rm{B}}}$ and ${F_{\rm{A}}}$ respectively into equation $\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ and making a the subject we obtain

$\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}$

Since ${m_{\rm{B}}}$ = 3kg and ${m_{\rm{B}}}$ = 7kg

g=9.81 and R=0.12m, I=0.22 ${\rm{ kg}} \cdot {{\rm{m}}^2}$

Substituting these we obtain

$a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}$

$\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the linear acceleration of block A is 1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(B)

For block B

${a_{\rm{B}}} = {a_{\rm{A}}}$

Therefore, the acceleration of both blocks A and B are same

1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(C)

The angular acceleration is $\alpha = \frac{a}{R}$

$\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

(D)

Tension on left side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}$

(E)

Tension on right side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}$

6 0

3 years ago

a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block

Airida [17]

Answer:

577g

Explanation:

Given parameters:

Temperature change = 5.9°C

Amount of heat lost = 427J

Unknown:

Mass of the block = ?

Solution:

The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.

H = m c Ф

H is the heat capacity

m is the mass of the block

c is the specific heat capacity

Ф is the temperature change

Specific heat capacity of lead is 0.126J/g°C

m = H / m Ф

m = $\frac{427}{0.126 x 5.9}$ = 577g

Mass of the lead block is 577g

7 0

3 years ago

Find equivalent resistance. <br><br>Answer asap and please, please don't spam.

Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

$\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega$

So, the equivalent resistance s 4.77 ohms.

4 0

3 years ago