Question

A person stands a distance R from a door's hinges and pushes with a force F directed perpendicular to its surface. By what factor does the applied torque change if the person's position and force change to the following? Use the definition of torque, with θ = 90°.(a) 2R and 2F(b) 2R and F(c) R and F/2(d) R/2 and F/2

Answer 1

Answer:

a)

4 times

b)

2 times

c)

0.5 times

d)

0.25 times

Explanation:

$F_{a}$ = Applied force by the person = F

$r$ = distance from the hinge = R

Torque is given as

$\tau = rF_{a}\\\tau = R F$

(a)

$F_{a}$ = Applied force by the person = 2F

$r$ = distance from the hinge = 2R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (2R) (2F)\\\tau_{new} = 4RF\\\tau_{new} = 4 \tau$

So Torque becomes 4 times

b)

$F_{a}$ = Applied force by the person = F

$r$ = distance from the hinge = 2R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (2R) (F)\\\tau_{new} = 2RF\\\tau_{new} = 2 \tau$

So Torque becomes 2 times

c)

$F_{a}$ = Applied force by the person = (0.5)F

$r$ = distance from the hinge = R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (R) (0.5)(F)\\\tau_{new} = (0.5)RF\\\tau_{new} = (0.5) \tau$

So Torque becomes 0.5 times

d)

$F_{a}$ = Applied force by the person = (0.5)F

$r$ = distance from the hinge = (0.5)R

New Torque is given as

$\tau_{new} = rF_{a}\\\tau_{new} = (0.5)(R) (0.5)(F)\\\tau_{new} = (0.25)RF\\\tau_{new} = (0.25) \tau$

So Torque becomes 0.25 times