Explanation:
a) Use the torque equation to solve for the amount of effort to lift the load:
[tex]\sum{\tau} = Lx_L - Ex_E = 0[/tex]
or
[tex]E = \dfrac{Lx_L}{x_E} = \dfrac{(100\:\text{N})(0.20\:\text{m})}{0.60\:\text{m}}[/tex]
[tex]\:\:\:\:\:=33.3\:\text{N}[/tex]
The mechanical advantage is
[tex]MA = \dfrac{0.60\:\text{m}}{0.20\:\text{m}} = 3[/tex]
The velocity ratio is the same as the MA:
[tex]VR = \dfrac{\text{effort arm}}{\text{load arm}} = MA = 3[/tex]
b) We can use the same equation in (a) to solve the problem:
[tex]\sum{\tau} = Ex_E - Lx_L = 0[/tex]
[tex]\Rightarrow x_L = \dfrac{Ex_E}{L} = \dfrac{(50\:\text {N})(90\:\text{cm})}{300\:\text{N}}[/tex]
[tex]\:\:\:\:\:=15\:\text{cm}[/tex]
c) We can write
[tex]W_RX_R = W_SX_S \Rightarrow X_R = \dfrac{W_SX_S}{W_R}[/tex]
[tex]\:\:\:\:\:= \dfrac{(400\:\text{N})(4\:\text{m})}{500\:\text{N}} = 3.2\:\text{m}[/tex]
d) We can solve the problem as follows:
[tex]\sum{\tau} = Ex_E - Lx_L = 0 \Rightarrow E = \dfrac{Lx_L}{x_E}[/tex]
or
[tex]E = \dfrac{(500\:\text{N})(0.50\:\text{m}}{1.0\:\text{m}} =250\:\text{N}[/tex]
The mechanical advantage MA is
[tex]MA = \dfrac{x_E}{x_L} = \dfrac{1.0\:\text{m}}{0.5\:\text{m}} = 2.0[/tex]
[tex]VR = MA = 2.0[/tex]
[tex]\%\text{eff} = \dfrac{Lx_L}{Ex_E}×100\%[/tex]
[tex]\:\:\:\:\:= \dfrac{(500\:\text{N})(0.5\:\text{m})}{(250\:\text{N})(1.0\:\text{m})}×100\%[/tex]
[tex]\:\:\:\:\:=100\%[/tex]
