Question

1. The three ropes in the figure are tied to a small, very light ring. Two of the ropes are anchored to walls at right angles, and the third rope pulls as shown. What are T1and T2, the magnitudes of the tension forces in the first two ropes?

1. The three ropes in the figure are tied to a small, very light ring. Two of the ropes are anchored to walls at right angles, and the third rope pulls as shown. What are T1and T2, the magnitudes of the tension forces in the first two ropes? A. Find the value of ax, the x-component of the object's acceleration. B. Find the value of ay, the y-component of the object's acceleration. A. What is the largest mass the ropes can support? 3. The forces in the figure are acting on a 3.3kg object. A. T1= B T2= 3. The two angled ropes used to support the crate in the figure below can withstand a maximum tension of 1700N before they break.

A. T1=

B T2=

3. The two angled ropes used to support the crate in the figure below can withstand a maximum tension of 1700N before they break.

A. What is the largest mass the ropes can support?

3. The forces in the figure are acting on a 3.3kg object.

A. Find the value of ax, the x-component of the object's acceleration.

B. Find the value of ay, the y-component of the object's acceleration.

Answer 1

Concepts and reason

The required concepts to solve the problem are tension, force and vectors.

First, form an equation for force in the vertical and horizontal direction. Solving these equations will give the tensions in the ropes.

By forming the equation for force in the horizontal direction, find the tension in the second rope. Substituting this tension in the equation of force in the vertical direction, the largest mass the ropes can support can be found.

From the given forces, find the angle made by the unknown force. Then find the unknown force using the given forces. By forming the equation of force in the vertical and horizontal direction, the accelerations can be found.

Fundamentals

A vector is a physical quantity with both magnitude and direction. The main examples of vectors are force and velocity. A vector can be resolved into components. If the angle made by the vector is with respect to the horizontal, then the horizontal component of the vector is the $\cos \theta$ component and the vertical component of the vector is the $\sin \theta$ component.

Force is the product of mass and acceleration. Tension is the state of being stretched tight. Force is given as

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

The resultant of two vectors is given as

$\vec v = \sqrt {v_x^2 + v_y^2}$

(A)

Along the horizontal direction, the net force is,

$\Sigma {F_x} = \left( {100\,{\rm{N}}} \right)\cos 30^\circ - {T_1}$

Here, ${F_x}$ is the force in the horizontal direction and ${T_1}$ is the tension of rope 1.

As there is no net force in the horizontal direction, the force is zero.

$0 = \left( {100\,{\rm{N}}} \right)\cos 30^\circ - {T_1}$

So, the tension in the first rope is,

$\begin{array}{c}\\{T_1} = \left( {100\,{\rm{N}}} \right)\cos 30^\circ \\\\ = 86.6\,{\rm{N}}\\\end{array}$

(B)

Along the vertical direction, the net force is,

$\Sigma {F_y} = {T_2} - \left( {100\,{\rm{N}}} \right)\sin 30^\circ$

Here, ${F_y}$ is the force in the horizontal direction and ${T_2}$ is the tension of second rope.

As there is no net force in the vertical direction, the force is zero.

$0 = {T_2} - \left( {100\,{\rm{N}}} \right)\sin 30^\circ$

So, the tension in the first rope is,

$\begin{array}{c}\\{T_2} = \left( {100\,{\rm{N}}} \right)\sin 30^\circ \\\\ = 50\,{\rm{N}}\\\end{array}$

(3)

The net horizontal force exerted on the system of rope is,

$\Sigma {F_x} = {T_1}\cos 45^\circ - {T_2}\cos 30^\circ$

Here, ${F_x}$ is the force in the horizontal direction, ${T_1}$ is the tension of the first rope and ${T_2}$ is the tension of second rope. There is no net force in the horizontal direction.

Then, the equation becomes

$0 = {T_1}\cos 45^\circ - {T_2}\cos 30^\circ$

Substitute $1700\,{\rm{N}}$ for ${T_1}$ to find ${T_2}$ .

$\begin{array}{c}\\{T_2} = \frac{{\left( {1700\,{\rm{N}}} \right)\cos 45^\circ }}{{\cos 30^\circ }}\\\\ = 1388.0{\rm{N}}\\\end{array}$

Along the vertical direction, the net force is,

$\Sigma {F_y} = {T_1}\sin 45^\circ + {T_2}\sin 30^\circ - mg$

Here, ${F_y}$ is the force in the vertical direction, ${T_1}$ is the tension of the first rope and ${T_2}$ is the tension of second rope. There is no net force in the vertical direction.

$0 = {T_1}\sin 45^\circ + {T_2}\sin 30^\circ - mg$

Now, the equation of mass becomes,

$m = \frac{{{T_1}\sin 45^\circ + {T_2}\sin 30^\circ }}{g}$

Substitute $1700\,{\rm{N}}$ for ${T_1}$ , $1388\,{\rm{N}}$ for ${T_2}$ and $9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $g$ to find $m$ .

$\begin{array}{c}\\m = \frac{{\left( {1700\,{\rm{N}}} \right)\sin 45^\circ + \left( {1388\,{\rm{N}}} \right)\sin 30^\circ }}{{9.8\,{\rm{m/}}{{\rm{s}}^2}}}\\\\ = 193.5\,{\rm{kg}}\\\end{array}$

(3.A)

The angle made by the unknown force is,

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{3.0\,{\rm{N}}}}{{4.0\,{\rm{N}}}}} \right)\\\\ = 36.87^\circ \\\end{array}$
‎

The unknown force can be found using the equation,

$\begin{array}{c}\\F = \sqrt {{{\left( {3.0\,{\rm{N}}} \right)}^2} + {{\left( {4.0\,{\rm{N}}} \right)}^2}} \\\\ = 5.0\,{\rm{N}}\\\end{array}$

The net force in the horizontal direction is,

$\begin{array}{l}\\\Sigma {F_x} = \left( {5.0\,{\rm{N}}} \right)\left( {\cos 36.87^\circ } \right) - \left( {2.0\,{\rm{N}}} \right)\\\\m{a_x} = \left( {5.0\,{\rm{N}}} \right)\left( {\cos 36.87^\circ } \right) - \left( {2.0\,{\rm{N}}} \right)\\\end{array}$

Here, $m$ is the mass of the object and ${a_x}$ is the horizontal acceleration.

The horizontal acceleration is,

${a_x} = \frac{{\left( {5.0\,{\rm{N}}} \right)\left( {\cos 36.87^\circ } \right) - \left( {2.0\,{\rm{N}}} \right)}}{m}$

Substitute $3.3\,{\rm{kg}}$ for $m$ to find ${a_x}$ .

$\begin{array}{c}\\{a_x} = \frac{{\left( {5.0\,{\rm{N}}} \right)\left( {\cos 36.87^\circ } \right) - \left( {2.0\,{\rm{N}}} \right)}}{{{\rm{3}}{\rm{.3}}\,{\rm{kg}}}}\\\\ = 0.606\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

(3.B)

The net force in the vertical direction is,

$\begin{array}{l}\\\Sigma {F_y} = \left( {5.0\,{\rm{N}}} \right)\left( {\sin 36.87^\circ } \right) - \left( {3.0\,{\rm{N}}} \right)\\\\m{a_y} = \left( {5.0\,{\rm{N}}} \right)\left( {\sin 36.87^\circ } \right) - \left( {3.0\,{\rm{N}}} \right)\\\end{array}$

Here, $m$ is the mass of the object and ${a_y}$ is the vertical acceleration.

The vertical acceleration is,

${a_y} = \frac{{\left( {5.0\,{\rm{N}}} \right)\left( {\sin 36.87^\circ } \right) - \left( {2.0\,{\rm{N}}} \right)}}{m}$

Substitute $3.3\,{\rm{kg}}$ for $m$ to find ${a_y}$ .

$\begin{array}{c}\\{a_y} = \frac{{\left( {5.0\,{\rm{N}}} \right)\left( {\sin 36.87^\circ } \right) - \left( {3.0\,{\rm{N}}} \right)}}{{{\rm{3}}{\rm{.3}}\,{\rm{kg}}}}\\\\ = 0{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part A

The magnitude of the tension force in the first rope is $86.6\,{\rm{N}}$ .

Part B

The magnitude of the tension force in the second rope is $50\,{\rm{N}}$ .

Part 3

The largest mass the ropes can support is $193.5\,{\rm{kg}}$ .

Part 3.A

The $x$ component of the object’s acceleration ${a_x}$ is $0.606\,{\rm{m/}}{{\rm{s}}^2}$ .

Part 3.B

The $y$ component of the object’s acceleration ${a_y}$ is $0\,{\rm{m/}}{{\rm{s}}^2}$ .