a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat

Question

Agata [3.3K]

3 years ago

15

a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat

e it uniformly? lift it 0.6 m from one end, pivoting around the opposite end

Physics

1 answer:

lukranit [14] · Answer 1 · 2022-07-20T07:40:48+03:00

1) The minimum force needed is 920 N

2) The minimum force is 460 N

3) The minimum force is 598 N

Explanation:

1)

We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:

Its weight, W, acting downward
The force applied to lift it, F, upward

So the net force on the railroad is

$\sum F = F - W$

where

W = 920 N is the weight of the railroad

In order to lift the railroad, the net force must be upward, so

$\sum F \geq 0$

And therefore

$F\geq W$

which means that the minimum force needed is equal to the weight of the railroad, 920 N.

2)

In this case, we have to use the principle of equilibrium of moments.

In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:

$W \frac{L}{2} = F L$

where

W = 920 N is the weight

L = 2.6 m is the length of the railroad

F is the force applied

We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.

Solving for F,

$F=\frac{W}{2}=\frac{920}{2}=460 N$

3)

This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.

This means that the distance between the point of application of the force F and the pivot is

L' = L - 0.6

where

L = 2.6 m

Therefore the equation for the equilibrium of moments becomes

$W\frac{L}{2}=F(L-0.6)$