Question

Please tell me answer please otherwise teacher will beat me fast ​

Answer 1

Explanation:

a) Use the torque equation to solve for the amount of effort to lift the load:

$\sum{\tau} = Lx_L - Ex_E = 0$

or

$E = \dfrac{Lx_L}{x_E} = \dfrac{(100\:\text{N})(0.20\:\text{m})}{0.60\:\text{m}}$

$\:\:\:\:\:=33.3\:\text{N}$

The mechanical advantage is

$MA = \dfrac{0.60\:\text{m}}{0.20\:\text{m}} = 3$

The velocity ratio is the same as the MA:

$VR = \dfrac{\text{effort arm}}{\text{load arm}} = MA = 3$

b) We can use the same equation in (a) to solve the problem:

$\sum{\tau} = Ex_E - Lx_L = 0$

$\Rightarrow x_L = \dfrac{Ex_E}{L} = \dfrac{(50\:\text {N})(90\:\text{cm})}{300\:\text{N}}$

$\:\:\:\:\:=15\:\text{cm}$

c) We can write

$W_RX_R = W_SX_S \Rightarrow X_R = \dfrac{W_SX_S}{W_R}$

$\:\:\:\:\:= \dfrac{(400\:\text{N})(4\:\text{m})}{500\:\text{N}} = 3.2\:\text{m}$

d) We can solve the problem as follows:

$\sum{\tau} = Ex_E - Lx_L = 0 \Rightarrow E = \dfrac{Lx_L}{x_E}$

or

$E = \dfrac{(500\:\text{N})(0.50\:\text{m}}{1.0\:\text{m}} =250\:\text{N}$

The mechanical advantage MA is

$MA = \dfrac{x_E}{x_L} = \dfrac{1.0\:\text{m}}{0.5\:\text{m}} = 2.0$

$VR = MA = 2.0$

$\%\text{eff} = \dfrac{Lx_L}{Ex_E}×100\%$

$\:\:\:\:\:= \dfrac{(500\:\text{N})(0.5\:\text{m})}{(250\:\text{N})(1.0\:\text{m})}×100\%$

$\:\:\:\:\:=100\%$