# Tension of a rope

1. Dec 5, 2017

### NoahCygnus

I think I messed up by not including the frictional force on the block + segment of rope system. Let me try again. So the system consists of the block and the rope A. Total mass of the system is $M_{system} = 13.3 kg$. The external forces on the system are friction and the tension.

Using Newton's second law:

$\Sigma F = ma$
$T_{1/2} - f_{kinetic} = M_{system}a$
$T_{1/2} = M_{system}a + f_{kinetic}$
$T_{1/2} = (13.3 kg)(1.3 ms^{-2}) + (0.5)(126) = 80.29 N$

Please correct me if I am wrong.

Last edited: Dec 5, 2017
2. Dec 5, 2017

### Abhishek kumar

You have taken g=10m/s^2 that's why you are getting more than 80N put g=9.8 result will be 79.1 N

3. Dec 6, 2017

### haruspex

Here are the results for five different views of the problem:
1. the simple view, in which the weight of the rope (but not its inertial mass) is ignored
2. the rod view, so half the rope's weight contributes to frictional force
3. the rope hangs down from the block but is horizontal at the 80N end; all of the rope's weight contributes to friction; the midpoint tension has an x and a y component
4. the rope hangs down at each end symmetrically (sort of.. the horizontal tension varies a bit along its length); the horizontal component of the applied 80N is reduced because some of that force is now vertical; half the rope's weight contributes to friction
5. the rope is horizontal at the block; the horizontal component of the 80N is reduced further as it takes the whole weight of the rope, while the midpoint tension again has a vertical component
Code (Text):

79.087 T, simple view

79.259 T, rod view

79.430 right hor, Tx
06.860 right hor, Ty
79.726 right hor, T

79.705 mid hor, Fx
78.979 mid hor, T=Tx

78.815 left hor, Fx
77.961 left hor, Tx
06.860 left hor, Ty
78.262 left hor, T

4. Dec 7, 2017

### Staff: Mentor

This is not exactly what I had in mind. What I was envisioning was the following:

If we do a differential force balance on the increment of rope between s and s + ds, we obtain:
$$\frac{d(T\sin{\phi})}{ds}=\rho g$$
$$\frac{d(T\cos{\phi})}{ds}=\rho a$$where s is the distance along the rope contour measured from the left end, $\phi$ is the angle that the tangent to the rope contour makes with the horizontal at location s, T is the tension in the rope at location s, $\rho$ is the linear density of the rope, and a is the horizontal acceleration.

If we integrate these equations from location s to the right end of the rope at s = L, we obtain:
$$T\sin{\phi}=T_L\sin{\phi_L}-\rho g (L-s)\tag{1}$$
$$T\cos{\phi}=T_L\cos{\phi_L}-\rho a (L-s)\tag{2}$$where $T_L$ is the tension at s = L (80 N in our problem) and $\phi_L$ is the contour angle at s = L.
In addition, the shape of the rope is described by:
$$\frac{dy}{ds}=\sin{\phi}\tag{3}$$
$$\frac{dx}{ds}=\cos{\phi}\tag{4}$$
We can eliminate the tension T from our equations by dividing Eqn. 1 by Eqn. 2 to yield:
$$\tan{\phi}=\frac{(dy/ds)}{\sqrt{1-(dy/ds)^2}}=\frac{T_L\sin{\phi_L}-\rho a(L-s)}{T_L\cos{\phi_L}-\rho g(L-s)}\tag{5}$$
Next, solving for dy/ds and reducing the equation to dimensionless form yields:
$$\frac{dY}{dS}=\frac{\sin{\phi_L}-\xi_g(1-S)}{\sqrt{(\sin{\phi_L}-\xi_g(1-S))^2+(\cos{\phi_L}-\xi_a(1-S))^2}}\tag{6}$$where Y=y/L, S=s/L, and where the dimensionless constants $\xi_g$ and $\xi_a$ are given by:
$$\xi_g=\frac{\rho g L}{T_L}$$and$$\xi_a=\frac{\rho a L}{T_L}$$
If we integrate Eqn. 6 from S = 0 (the left end of the rope) to S = 1 (the right end of the rope), we obtain the difference between the dimensionless rope elevations at the two ends:
$$Y(1)-Y(0)=\int_0^1{\frac{\sin{\phi_L}-\xi_g(1-S)}{\sqrt{(\sin{\phi_L}-\xi_g(1-S))^2+(\cos{\phi_L}-\xi_a(1-S))^2}}dS}\tag{7}$$
If the two ends of the rope are at the same elevation, then we must have that:
$$\int_0^1{\frac{\sin{\phi_L}-\xi_g(1-S)}{\sqrt{(\sin{\phi_L}-\xi_g(1-S))^2+(\cos{\phi_L}-\xi_a(1-S))^2}}dS}=0\tag{7}$$This equation will be satisfied if the proper value is employed for the contour angle at the right hand end of the rope $\phi_L$. Thus, Eqn. 7 provides a relationship for determining $\phi_L$. Once $\phi_L$ is known, the value of the contour angle and tension at all locations along the rope are established, including the values at the left end where the cord is attached to the block.

5. Dec 7, 2017

### haruspex

Ah, you wanted the full catenary treatment. I was thinking just in terms of getting good approximations in the present context so that we could compare the consequences of the different feasible assumptions.
I believe the difference the catenary analysis would make is very small, since it depends principally on how the horizontal mass distribution of the rope differs from uniform. In the present problem that would be tiny.