Question

In each caselet be the weight of the suspended crate full of priceless artobjects. The strut is uniform and also has weight .

Part AFind the tension in the cable in the arrangement (a).Express your answer in terms of .

=

Part BFind the magnitude of the force exerted on the strut by the pivot in the arrangement (a).Express your answer in terms of .

=

Part CFind the direction of the force exerted on the strut by the pivot in the arrangement (a).

=	$^circ$ from thehorizontal

Part DFind the tension in the cable in the arrangement (b).Express your answer in terms of .

=

Part EFind the magnitude of the force exerted on the strut by the pivot in the arrangement (b).Express your answer in terms of .

=

Part FFind the direction of the force exerted on the strut by the pivot in the arrangement (b).

=		$^circ$ from thehorizontal

Answer 1

a) 

b) 

c) 

d)

e)

f)

Answer 2

Concepts and reason

The concepts that are to be used to solve the given problem are force, torque, and a trigonometric identity to find the direction of force.

Find the individual torques on the system and then use equilibrium of torque to find the tension in the cable.

Use the equilibrium of force along the horizontal and vertical directions and find the force exerted by the pivot point.

Use the trigonometric identify to find the direction of the force on the strut by the pivot point.

Fundamentals

The torque is defined as the product of force and the perpendicular distance of the force.

At equilibrium, the net torque on the system must be equal to zero.

(a)

The net torque on the strut is zero in equilibrium situation.

$\begin{array}{c}\\\sum \tau = 0\\\\T\left( {L\tan {{30}^{\rm{o}}}} \right) - wL - w\left( {\frac{L}{2}} \right) = 0\\\\T = 2.60w\\\end{array}$

Here, T is the tension in the cable and w is the weight of the crate and also the weight of the struct.

(b)

Apply equilibrium of force along the horizontal direction and find the horizontal component of force applied by the pivot point ${F_x}$ .

${F_x} = 2.60w$

Apply equilibrium of force along the vertical direction and find the vertical component of force applied by the pivot point ${F_y}$ .

${F_y} = 2w$

Now, use the following formula to find the magnitude of the force on the struct by the pivot point.

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}}$

Substitute $2.60w$ for ${F_x}$ and $2w$ for ${F_y}$ .

$\begin{array}{c}\\F = \sqrt {{{\left( {2.60w} \right)}^2} + {{\left( {2w} \right)}^2}} \\\\ = 3.28w\\\end{array}$

(c)

The trigonometric identity to find the direction of the force on the strut by the pivot is,

$\begin{array}{c}\\\tan \theta = \frac{{2w}}{{2.60w}}\\\\\theta = {37.6^{\rm{o}}}\\\end{array}$

[Part c]

Part c

(d)

The net torque on the strut is zero in equilibrium situation.

$\begin{array}{c}\\\sum \tau = 0\\\\T\left( {L\sin {{15}^{\rm{o}}}} \right) - wL\sin {45^{\rm{o}}} - \frac{w}{2}\left( {L\sin {{45}^{\rm{o}}}} \right) = 0\\\\T = 4.10w\\\end{array}$

(e)

Apply equilibrium of force along the horizontal direction and find the horizontal component of force applied by the pivot point ${F_x}$ .

$\begin{array}{c}\\{F_x} = T\cos {30^{\rm{o}}}\\\\ = \left( {4.10w} \right)\cos {30^{\rm{o}}}\\\\ = 3.55w\\\end{array}$

Apply equilibrium of force along the vertical direction and find the vertical component of force applied by the pivot point ${F_y}$ .

$\begin{array}{c}\\{F_y} = 2w + \left( {4.10w} \right)\sin {30^{\rm{o}}}\\\\ = 4.05w\\\end{array}$

Now, use the following formula to find the magnitude of the force on the struct by the pivot point.

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}}$

Substitute $3.55w$ for ${F_x}$ and $4.05w$ for ${F_y}$ .

$\begin{array}{c}\\F = \sqrt {{{\left( {3.55w} \right)}^2} + {{\left( {4.05w} \right)}^2}} \\\\ = 5.39w\\\end{array}$

(f)

The trigonometric identity to find the direction of the force on the strut by the pivot is,

$\tan \theta = \frac{{{F_y}}}{{{F_x}}}$

Substitute $3.55w$ for ${F_x}$ and $4.05w$ for ${F_y}$ .

$\begin{array}{c}\\\tan \theta = \frac{{4.05w}}{{3.55w}}\\\\\theta = {48.8^{\rm{o}}}\\\end{array}$

Ans: Part a

Thus, the tension in the cable is 2.60w.

Part b

Thus, the force on the struct by the pivot is $3.28w$ .

Part c

Thus, the direction of the force on the strut by the pivot is ${37.6^{\rm{o}}}$ .

Part d

Thus, the tension in the cable is 4.10w.

Part e

Thus, the force on the struct by the pivot is $5.39w$ .

Part f

Thus, the direction of the force on the strut by the pivot is ${48.8^{\rm{o}}}$ .