Answer:
![x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7BL%7D%7Btan%28%5Cphi_1%29cot%28%5Cphi_2%29%2B1%7D)
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)](https://tex.z-dn.net/?f=F_%7B1x%7D%20%3D%20-F_1cos%28%5Cphi_1%29%5C%5CF_%7B1y%7D%3DF_1sin%28%5Cphi_1%29%5C%5C%5C%5CF_%7B2x%7D%3DF_2cos%28%5Cphi_2%29%5C%5CF_%7B2y%7D%3DF_2sin%28%5Cphi_2%29)
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](https://tex.z-dn.net/?f=%5Csum%20F_x%3D0%5C%5CF_%7B1x%7D%3DF_%7B2x%7D%5C%5CF_1cos%28%5Cphi_1%29%3DF_2cos%28%5Cphi_2%29%5C%5C%5Cfrac%7BF_1%7D%7BF_2%7D%3D%5Cfrac%7Bcos%28%5Cphi_2%29%7D%7Bcos%28%5Cphi_1%29%7D-------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
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}}](https://tex.z-dn.net/?f=-F_%7B1y%7D%5Ctimes%20x%20%2B%20F_%7B2y%7D%28L-x%29%20%3D%200%3C%2Fp%3E%3Cp%3E%5C%5CF_%7B1y%7D%5Ctimes%20x%3DF_%7B2y%7D%28L-x%29%5C%5Cx%3D%5Cfrac%7BF_%7B2y%7D%28L-x%29%7D%7BF_%7B1y%7D%7D)
But, ![F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)](https://tex.z-dn.net/?f=F_%7B1y%7D%3DF_1sin%28%5Cphi_1%29%5C%20and%5C%20F_%7B2y%7D%3DF_2sin%28%5Cphi_2%29)
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\\\\](https://tex.z-dn.net/?f=x%3D%5Cfrac%7BF_2sin%28%5Cphi_2%29%28L-x%29%7D%7BF_1sin%28%5Cphi_1%29%7D%5C%5C%5Ctextrm%7BFrom%20equation%20%281%29%2C%7D%5Cfrac%7BF_2%7D%7BF_1%7D%3D%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5C%5C%5Ctherefore%20x%3D%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7D%5Ctimes%20%28L-x%29%5C%5Cx%3D%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7D%5Ctimes%20L-%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7D%5Ctimes%20x%5C%5C%5C%5C)
![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](https://tex.z-dn.net/?f=x%281%2B%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7D%29%3D%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7DL%5C%5Cx%281%2B%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bsin%28%5Cphi_1%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bcos%28%5Cphi_2%29%7D%29%3D%5Cfrac%7Bcos%28%5Cphi_1%29%7D%7Bcos%28%5Cphi_2%29%7D%5Ctimes%20%5Cfrac%7Bsin%28%5Cphi_2%7D%7Bsin%28%5Cphi_1%29%7DL)
We know,
![tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}](https://tex.z-dn.net/?f=tan%28%5Cphi%29%3D%5Cfrac%7Bsin%28%5Cphi%29%7D%7Bcos%28%5Cphi%29%7D%5C%5C%5C%5Ccot%28%5Cphi%29%3D%5Cfrac%7Bcos%28%5Cphi%29%7D%7Bsin%28%5Cphi%29%7D)
∴![x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7BL%7D%7Btan%28%5Cphi_1%29cot%28%5Cphi_2%29%2B1%7D)
Answer:
mixtures that appear to be the same throughout are homogenous mixtures
Answer:
m = 20,000 kg
Explanation:
Force, ![F=2.5\times 10^4\ N](https://tex.z-dn.net/?f=F%3D2.5%5Ctimes%2010%5E4%5C%20N)
Acceleration of the shark, ![a=1.25\ m/s^2](https://tex.z-dn.net/?f=a%3D1.25%5C%20m%2Fs%5E2)
It is required to find the mass of the shark. Let m is the mass. Using second law of motion to find it as follows :
F = ma
Putting the value of F and a to find m
![m=\dfrac{F}{a}\\\\m=\dfrac{2.5\times 10^4}{1.25}\\\\m=20,000\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BF%7D%7Ba%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B2.5%5Ctimes%2010%5E4%7D%7B1.25%7D%5C%5C%5C%5Cm%3D20%2C000%5C%20kg)
So, the shark's mass is 20,000 kg.
Answer:
I would say it's true
Explanation:
when u get older we have less bones