Question

An open-link chain of length L and mass $\rho$ per unit length is released from rest on the incline with x = 0. The coefficient of kinetic friction between the chain and the incline is $\mu_k$. The chain piles up in a bucket at the base of the incline. Determine as a function of x the added force R on the bucket required to hold it in place against the impact of the chain. Note that the last link in motion at the bottom of the incline has lost contact with the link which immediately preceded it.

Answer 1

$\rho g x (3 \sin\theta - 2 \mu_k \cos\theta)$

Answer 2

The problem asks for the added force R on the bucket required to hold it in place against the impact of the chain as a function of x, the length of the chain that has already entered the bucket.

1. Determine the acceleration of the moving part of the chain:

Let the length of the chain on the incline be $(L-x)$. The mass of this moving part is $m_{moving} = \rho(L-x)$.

The forces acting on this moving part along the incline are:

• Component of gravitational force down the incline: $F_g = m_{moving} g \sin\theta = \rho(L-x) g \sin\theta$.
• Kinetic friction force up the incline: $f_k = \mu_k N = \mu_k m_{moving} g \cos\theta = \mu_k \rho(L-x) g \cos\theta$.

The net force on the moving chain is:

$F_{net} = F_g - f_k = \rho(L-x) g \sin\theta - \mu_k \rho(L-x) g \cos\theta$

$F_{net} = \rho(L-x) g (\sin\theta - \mu_k \cos\theta)$

The acceleration of the chain, $a$, is given by $F_{net} = m_{moving} a$:

$a = \frac{\rho(L-x) g (\sin\theta - \mu_k \cos\theta)}{\rho(L-x)}$

$a = g (\sin\theta - \mu_k \cos\theta)$

Since $g$, $\sin\theta$, and $\mu_k \cos\theta$ are constants, the acceleration $a$ is constant. Let's denote it as $a_0$.

2. Determine the velocity of the chain links entering the bucket:

The chain is released from rest (initial velocity $u=0$). When a length $x$ of the chain has entered the bucket, the chain has moved a distance $x$ down the incline.

Using the kinematic equation $v^2 = u^2 + 2as$:

$v^2 = 0^2 + 2 a_0 x$

$v^2 = 2 g x (\sin\theta - \mu_k \cos\theta)$

3. Determine the total force on the bucket:

The force R required to hold the bucket in place must counteract two effects:

• The component of the weight of the chain already accumulated in the bucket along the incline.
• The impact force from the incoming chain links.

a. Weight of accumulated chain:

The mass of the chain already in the bucket is $m_{acc} = \rho x$.

The component of its weight acting down the incline is $W_{acc,x} = m_{acc} g \sin\theta = \rho x g \sin\theta$.

b. Impact force from incoming chain:

This is a variable mass system problem. The rate at which mass enters the bucket is $\frac{dm}{dt}$.

Since the chain is moving with velocity $v$, a length $dx$ enters the bucket in time $dt$, so $dx = v dt$.

The mass entering in time $dt$ is $dm = \rho dx = \rho v dt$.

Therefore, the rate of mass accumulation is $\frac{dm}{dt} = \rho v$.

The impact force $F_{impact}$ is given by the rate of change of momentum of the incoming mass. Since the bucket is held stationary, the relative velocity of the incoming mass with respect to the bucket is $v$.

$F_{impact} = v \frac{dm}{dt} = v (\rho v) = \rho v^2$. This force acts down the incline.

The total force R required to hold the bucket in place (acting up the incline, as shown in the diagram) is the sum of these two forces:

$R = W_{acc,x} + F_{impact}$

$R = \rho x g \sin\theta + \rho v^2$

4. Substitute the expression for $v^2$ into the equation for R:

$R = \rho x g \sin\theta + \rho [2 g x (\sin\theta - \mu_k \cos\theta)]$

$R = \rho x g \sin\theta + 2 \rho g x \sin\theta - 2 \rho g x \mu_k \cos\theta$

Combine the terms with $\sin\theta$:

$R = 3 \rho g x \sin\theta - 2 \rho g x \mu_k \cos\theta$

Factor out common terms $\rho g x$:

$R = \rho g x (3 \sin\theta - 2 \mu_k \cos\theta)$

The added force R on the bucket required to hold it in place against the impact of the chain, as a function of x, is:

$R = \rho g x (3 \sin\theta - 2 \mu_k \cos\theta)$

The problem statement "Note that the last link in motion at the bottom of the incline has lost contact with the link which immediately preceded it" simplifies the problem by allowing us to treat the incoming links as individual masses impacting the bucket, rather than a continuous chain where tension might play a role in the impact force calculation.