Question

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with the horizontal, and the right wire makes an angle ϕ2. The bar has length L.

Answer 1

Answer:

$x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

Explanation:

Let 'F₁'  and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.

Now, the horizontal and vertical components of these forces are:

$F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)$

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

$\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1$

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces $F_{1y}\ and\ F_{2y}$ will be zero. This gives,

$-F_{1y}\times x + F_{2y}(L-x) = 0

\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}$

But, $F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)$

Therefore,

$x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x$

$x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L$

We know,

$tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}$

∴$x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

Answer 2

The position of the center of mass of the bar measured as a distance x  from the bar's left :  

 $X = \frac{L *tan \alpha /2 }{tan\alpha _{1} + tan\alpha _{2} }$

Determine the position of the center of mass

First step : State the static equation conditions

Σfx = 0

T₁cos$\alpha _{1}$ = T₂cos$\alpha _{2}$ ------- ( 1 )

Σfy = 0

T₁sin$\alpha _{1}$ = T₂sin$\alpha _{2}$ ------- ( 2 )

Σw ( taking moment about w ) = 0

T₁sin$\alpha _{1}$ ( x ) = T₂sin$\alpha _{2}$ ( 1 - x ) ------- ( 3 )

T₁sin$\alpha _{1}$ = T₂sin$\alpha _{2}$ ( 1 - x ) / x

Next step : Determine the position of the center of mass

Back to equation ( 2 )

T₂sin$\alpha _{2}$ =  wx / L

T₁sin$\alpha _{1}$ = w ( L-x ) / L

Back to equation ( 1 )

[ w ( L- x ) / L sin $\alpha$₁ ]  cos $\alpha$₁  =  [ wx / Lsin$\alpha _{2}$ ] cos$\alpha _{2}$

( L - x ) tan$\beta _{2}$  = x tan $\alpha _{1}$  ----- ( 4 )

Solve for X  

$X = \frac{L *tan \alpha /2 }{tan\alpha _{1} + tan\alpha _{2} }$

Hence we can conclude The position of the center of mass of the bar measured as a distance x  from the bar's left :  

 $X = \frac{L *tan \alpha /2 }{tan\alpha _{1} + tan\alpha _{2} }$

Learn more about center of mass : https://brainly.com/question/874205

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