Answer to Question #164850 in Mechanics | Relativity for Imoh

Question

A bottle opener requires that a force of 35N must be applied to the handle in order to lift the bottle cap 0.90m. the opener has a V.R of 8.0 and an efficiency of 75%. 1. What is the M.A of the opener

2. What force is applied to the bottle cap

3. How far does the handle of the opener move

Answer 1

Given,

Force required, F=35N

Displacement of handle, $d_l=0.90m$

V.R.=8, $\eta=0.75$ ,

(1) Mechanical Advantage, $M.A.=\eta\times V.R.$

$=0.75\times8=6$

(2) Forced applied or bottle cap

$F_e=\dfrac{F_l}{MA}$

$=\dfrac{35}{6}=5.83N$

(3) As the bottle cap is in equilibrium so

net torque is zero.

$f_l\times d_l=f_e\times d_e$

$35\times d_l=5.83\times 0.9$

$d_l=\dfrac{5.247}{35}=0.1499\equiv 0.15$

Hence the handle move 0.15m