Calculate Forces at A & B in Steel Wire Tractor Pull

[raymanmusic]

Homework Statement

https://dl.dropboxusercontent.com/u/11241083/tractor.png
How do I calculate reaction forces at A and B? We pull with a force F2 = 300 N on the steel wire. The steel wire goes two and a half times around the axel wheel.

$d = 38 mm$

$D = 200 mm$

$a = 100 mm$

$b = 120 mm$

$\sigma_{w} = 256 \frac{N}{mm^{2}}$

$K_{t} = 1.5$

$\eta_{k}=0.7$

$\mu=0.2$

Correct answers:
$F_{A} = 13.3 kN$
$F_{B} = 6 kN$

 

[SteamKing]

What have you tried? This seems to be a very straight forward problem.

Why don't you try something that can be checked and discussed?

Hint: Pretend the power take of shaft (a-b-d) is a beam with two supports and the load from the sheave is cantilevered over the end of this beam.

 

[raymanmusic]

What is the force at the end of the shaft? Is it the same as F2=300N? I sum moment at point B to try to find reaction force at A.
https://dl.dropboxusercontent.com/u/11241083/New%20Document20140407191616593.pdf

 

[SteamKing]

What is the force at the end of the shaft? Is it the same as F2=300N? I sum moment at point B to try to find reaction force at A.
https://dl.dropboxusercontent.com/u/11241083/New%20Document20140407191616593.pdf

Look at the left figure, the one showing how the line goes around the sheave. You are given F2, so make a FBD of the sheave to see what load the shaft sees.

 

[rezeew]

To calculate the reaction forces at points A and B, we need to use the equations for static equilibrium. This means that the sum of all forces acting on the system must be equal to zero, and the sum of all moments must also be equal to zero.

First, we need to draw a free body diagram of the system, showing all the forces acting on it. In this case, we have the force F2 acting on the steel wire, the reaction forces F_A and F_B at points A and B, respectively, and the weight of the tractor pulling on the wire.

Next, we can write out the equations for static equilibrium:

ΣF_x = 0 (sum of all forces in the x-direction)
ΣF_y = 0 (sum of all forces in the y-direction)
ΣM = 0 (sum of all moments)

We can start by looking at the x-direction. We know that the only force acting in the x-direction is F_A, so we can write:

ΣF_x = F_A = 0

This means that F_A must be equal to zero, as there are no other forces acting in the x-direction.

Next, we can look at the y-direction. We have F2 pulling in the positive y-direction, and F_B acting in the negative y-direction. We also have the weight of the tractor acting downward. So we can write:

ΣF_y = F2 - F_B - W = 0

We know that F2 = 300 N, and we can calculate the weight of the tractor using the formula W = mg, where m is the mass of the tractor and g is the acceleration due to gravity. We can assume a mass of 1000 kg for the tractor, so W = 1000 kg x 9.8 m/s^2 = 9800 N.

Plugging in these values, we get:

300 N - F_B - 9800 N = 0
F_B = 9800 N - 300 N = 9500 N

Now, we can move on to calculating the moment at point A. We know that the force F2 is acting at a distance of 2.5D from point A, and the weight of the tractor is acting at a distance of (a+b) from point A. So we can write:

ΣM = F2 x 2.5D - W x (a