Answer:
2.5 m
Explanation:
Weight of billboard worker = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension in rope = 550 N
The sum of the torques will be
![-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m](https://tex.z-dn.net/?f=-800%284-x%29-500%5Ctimes%202%2B550%5Ctimes%204%3D0%5C%5C%5CRightarrow%20-800%284-x%29%3D500%5Ctimes%202-550%5Ctimes%204%5C%5C%5CRightarrow%20-800%284-x%29%3D-1200%5C%5C%5CRightarrow%20-x%3D%5Cdfrac%7B1200%7D%7B800%7D-4%5C%5C%5CRightarrow%20-x%3D-2.5%5C%5C%5CRightarrow%20x%3D2.5%5C%20m)
The position of the person will be 2.5 m
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book![(m_0)=1.7 kg](https://tex.z-dn.net/?f=%28m_0%29%3D1.7%20kg)
height(h)=1 m
time taken=0.64 s
![h=ut+frac{at^2}{2}](https://tex.z-dn.net/?f=h%3Dut%2Bfrac%7Bat%5E2%7D%7B2%7D)
![1=0+\frac{a(0.64)^2}{2}](https://tex.z-dn.net/?f=1%3D0%2B%5Cfrac%7Ba%280.64%29%5E2%7D%7B2%7D)
![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley
![4.88=\alpha \times 5.2\times 10^{-2}](https://tex.z-dn.net/?f=4.88%3D%5Calpha%20%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D)
![\alpha =93.84 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D93.84%20rad%2Fs%5E2)
And Tension in Rope
![T=m(g-a)](https://tex.z-dn.net/?f=T%3Dm%28g-a%29)
![T=1.7\times (9.8-4.88)](https://tex.z-dn.net/?f=T%3D1.7%5Ctimes%20%289.8-4.88%29)
T=8.364 N
and Tension will provide Torque
![T\times r=I\cdot \alpha](https://tex.z-dn.net/?f=T%5Ctimes%20r%3DI%5Ccdot%20%5Calpha%20)
![8.364\times 5.2\times 10^{-2}=I\times 93.84](https://tex.z-dn.net/?f=8.364%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D%3DI%5Ctimes%2093.84)
![I=0.463\times 10^{-2} kg-m^2](https://tex.z-dn.net/?f=I%3D0.463%5Ctimes%2010%5E%7B-2%7D%20kg-m%5E2)
![I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2](https://tex.z-dn.net/?f=I_%7Boriginal%7D%3D%5Cfrac%7Bmr%5E2%7D%7B2%7D%3D0.31%5Ctimes%2010%5E%7B-2%7Dkg-m%5E2)
Thus mass is uniformly distributed or some more towards periphery of Pulley
A-200 kg
I hope this helped xx
Answer:
5.38 m/s^2
Explanation:
NET force causing the object to accelerate = 50 -10 = 40 N
Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg
F = ma
40 = 7.44 * a a = 5.38 m/s^2