Question

An overhead garage door has two springs, one on each side of the door. A force of 15 pounds is required to stretch each spring 1 foot. Because of a pulley system, the springs stretch only one-half the distance the door travels. The door moves a total of 8 feet, and the springs are at their natural lengths when the door is open. Find the combined lifting force applied to the door by the springs when the door is closed.

Answer 1

Answer:

120 pounds

Step-by-step explanation:

Data provided in the question:

Force applied, F₁ = 15 pounds

Displacement by Force F₁, x₁ = 1 foot

At natural length of the spring total movement of the door = 8 feet

Now,

we know,

Force in spring, F = kx

where, k is the spring constant

Thus,

15 = k(1)      ............(1)

now when the door travels half the distance x₂ = $\frac{8}{2}$ = 4 feet

Therefore,

F₂ = k(4)      ..........(2)

Dividing 2 by 1

we get

$\frac{F_2}{F_1}=\frac{k(4)}{k(1)}$

or

$\frac{F_2}{15}=\frac{(4)}{(1)}$

or

F = 15 × 4 = 60 pounds

Thus,

for 2 springs combined force applied = 2 × 60 pounds = 120 pounds