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

Answer 2

The factor by which the torque changes at the given forces and distance are;

A) Torque changes by a factor of 4.

B) Torque changes by a factor of 2.

C) Torque changes by a factor of ¹/₂.

D) Torque changes by a factor of ¹/₄.

Formula for torque in this question is;

Torque = Force × distance

Thus;

τ = FR

A) When Distance and Force are respectively; 2R and 2F;

τ = 2F × 2R

τ = 4(FR)

So torque changes by a factor of 4.

B) When Distance and Force are respectively; 2R and F;

τ = F × 2R

τ = 2(FR)

So torque changes by a factor of 2.

C) When Distance and Force are respectively; R and F/2;

τ = (F/2) × (R)

τ = ¹/₂(FR)

So torque changes by a factor of ¹/₂.

D) When Distance and Force are respectively; R/2 and F/2;

τ = (F/2) × (R/2)

τ = ¹/₄(FR)

So torque changes by a factor of ¹/₄.

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