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Llana [10]
3 years ago
15

A car is being towed at a constant velocity on a horizontal road using a horizontal chain. The tension in the chain must be equa

l to the weight of the car in order to maintain a constant velocity.
True or False?
Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

<em>The statement is false</em>

Explanation:

<u>Dynamics</u>

To analyze the situation stated in the question, we must apply some basic knowledge about dynamics, specifically Newton's laws that explain how acceleration is produced or not, depending on the forces acting on a system of particles.

The second Newton's law states the acceleration of an object of mass m can be found by knowing the net force Fn applied to it with the formula

F_n=m.a

Considering the kinetics equations, we know that

\displaystyle a=\frac{v_f-v_o}{t}

Where vf-vo is the change of velocity in time t. If an object has zero acceleration, it means its velocity doesn't change, it has constant velocity.

Under such conditions, then the net force is also zero.

We know the car is being towed horizontally at a constant velocity, meaning it has a net force equal to zero in the horizontal direction. In other words, all the horizontal forces are balanced.

We also know the chain is applying a horizontal force to tow the car, so there must be another force opposing tho that force making the car moving at a constant velocity. Although it's not mentioned, the friction force must be acting to stop the car and compensating the tension T exerted by the chain.

The friction force is given by

F_r=\mu N

Where \mu is the friction coefficient and N is the normal force, which is equal to the weight of the object W at the conditions of the problem.

Knowing the net force is zero, then

\mu W=T

If the tension is equal to the weight of the car, then

\mu T=T

simplifying

\mu=1

Thus, the only way the tension and the weight are equal is when the friction coefficient is 1, so the initial assumption is false

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two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of
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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
Explain in detail what specific factors affect the momentum of a bicycle going down hill.
Crazy boy [7]

Mass and velocity are the two terms which affect momentum of a bicycle going hill down.

Explanation:

As we know that Momentum describes the motion of an object. It is the combination of the objects mass and velocity.

So, obviously with no doubt mass and velocity are the two terms which affect momentum.

Momentum(p) = Mass(m) * Velocity(v)

The momentum also depends upon the mass and speed of the object.

More the mass of the object more is the momentum.

Depending upon the gravity and bicycle's motion speed momentum varies.

Bicycle moves faster the down hill if it moves with some speed as it has lesser mass the momentum also will be less.

4 0
3 years ago
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