Chapter 16.1, Problem 16.39P

A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are ${\mu }_s=0.50$ and ${\mu }_k=0.40$ . For $P=\text{3}.6\text{lb}$ , determine (a) whether slipping occurs between the belt and either cylinder, (b) the angular acceleration of each cylinder.

To determine

i.

If there is any slipping between the belt and the cylinders and the angular acceleration of two cylinders.

Answer to Problem 16.39P

There is slipping between cylinder b and belt.

Angular acceleration of cylinder a = ${\alpha }_{\text{B}}\text{ = 9}{\text{.659 rad/s}}^2$

Angular acceleration of cylinder A = ${\text{α}}_A\text{ = 61}{\text{.824 rad/s}}^2$

Explanation of Solution

Given:

weight of Cylinder A, w_a = 5 lb radius of cylinder A = 4 in.

weight of cylinder B, w_b = 20 lb

radius of cylinder B, r_B = 8 in.

Magnitude of Force pulling the belt = 3.6 lb

Coefficient of static friction = 0.5

Coefficient of kinetic friction = 0.4

Concept used:

Condition of non-slipping,

Friction force due to static friction must be greater than the pulling force on each cylinder.

F_s> F_A, F_s>F_B

$\begin{array}{l}{\text{F}}_s{\text{ = μ}}_{\text{s}}{\text{N> F}}_{\text{A}}\\ {\text{F}}_s{\text{ = μ}}_{\text{s}}{\text{N> F}}_{\text{B}}\\ {\text{ In case of slipping, maximum tangential force between a cylinder and belt is F}}_s\text{ , }\\ {\text{otherwise it F}}_A{\text{ and F}}_B\text{, respectively for cylinder A and B}\end{array}$

Mass moment of inertia for a disk is given by-

$\text{I = }\frac{1}{2}{\text{mr}}^2,\text{ where r is the radius of disk }$

The tangential force acting on a cylinder will provide the angular acceleration to the cylinder. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα }\end{array}$

Calculation:

Mass of cylinder A = $\frac{\text{5lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.15528 lb-s}}^{\text{2}}\text{/ft}$

Mass of cylinder B = $\frac{\text{20lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.62112 lb-s}}^{\text{2}}\text{/ft}$

Mass moment of inertia of cylinder A = $\frac{1}{2}{\text{m}}_{\text{A}}{r}_{\text{A}}^{\text{2}}\text{ = }\frac{1}{2}\text{(0}{\text{.15528 lb-s}}^{\text{2}}\text{/ft)×(}\frac{4}{\text{12}}{\text{)}}^{\text{2}}\text{ = 8}{\text{.6267×10}}^{\text{-3}}{\text{lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of cylinder B = $\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}\text{ = }\frac{1}{2}\text{(0}{\text{.62112 lb-s}}^{\text{2}}{\text{/ft)×(8/12)}}^{\text{2}}\text{ = 0}{\text{.13803 lb×s}}^{\text{2}}\text{ft}$

$\begin{array}{l}\text{tangential acceleration of cylinder A is equal to the tangential acceleration of cylinder B}\\ \text{The relation between acceleration and angular acceleration}\\ \text{a = rα}\\ {\text{Therefore, a}}_{\text{t}}{\text{ = r}}_{\text{A}}{\text{α}}_{\text{A}}{\text{ = r}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{4α}}_{\text{A}}{\text{ = 8α}}_{\text{B}}\\ {\text{α}}_{\text{A}}{\text{ = 2α}}_{\text{B}}\end{array}$

For cylinder A,

$\begin{array}{l}{\text{ΣM}}_{\text{A}}{\text{ = I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{F}}_{\text{A}}{\text{×r}}_{\text{A}}{\text{= I}}_{\text{A}}{\text{α}}_{\text{A}}\\ {\text{F}}_{\text{A}}\text{×(4/12 ft) = (8}{\text{.6267×10}}^{\text{-3}}{\text{) lb×ft×s}}^{\text{2}}{\text{ × (2α}}_B\text{)}\\ {\text{F}}_{\text{A}}\text{= 0}{\text{.05176α}}_B\text{ lb}\end{array}$

For cylinder B,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{B}}{\text{×r}}_B{\text{= I}}_{\text{B}}{\text{α}}_{\text{B}}\end{array}$

$\begin{array}{l}{\text{F}}_{\text{B}}\text{×(8/12 ft) = (0}{\text{.13803) lb×ft×s}}^{\text{2}}{\text{ × (α}}_B\text{)}\\ {\text{F}}_{\text{B}}\text{ = 0}{\text{.20705α}}_B\end{array}$

the net force on the belt should be summation of P, force by cylinder B and force by cylinder C, all equals to 0.

$\begin{array}{l}{\text{ΣF = P- F}}_{\text{A}}{\text{-F}}_B\text{ = 0}\\ \text{3}{\text{.6 lb - F}}_{\text{A}}{\text{-F}}_{\text{B}}\text{ = 0 }\\ \text{3}\text{.6lb - 0}{\text{.05176α}}_{\text{B}}\text{ - 0}{\text{.20705α}}_{\text{B}}\text{ = 0}\\ {\text{α}}_{\text{B}}\text{ = 13}{\text{.91 rad/s}}^{\text{2}}\\ {\text{α}}_A\text{ = 27}{\text{.8196 rad/s}}^{\text{2}}\\ {\text{F}}_A\text{ = 0}{\text{.05176 α}}_{\text{B}}\text{ = 0}\text{.72 lb}\\ {\text{F}}_B\text{ = 0}{\text{.20705α}}_{\text{B}}\text{ = 2}\text{.88 lb }\end{array}$

Maximum friction force to avoid slipping F_s = ${\text{μ}}_{\text{s}}\text{N = 0}\text{.5}\times \text{5lb = 2}\text{.5 lb}$

Condition of slipping is ${\mu }_{s}N>{\text{ F}}_B$

$\begin{array}{l}{\mu }_{s}N`=\text{ 2}\text{.5 lb,}\\ {\text{F}}_B\text{ = 2}\text{.88 lb}\\ \text{The condition of non-slipping is not satisfied}\text{. Therefore slipping occurs between the belt and the cylinder B}\end{array}$

Conclusion:

There is slipping between cylinder b and belt.

To determine

ii.

As there sliping, find the angular acceleration of two cylinders.

Answer to Problem 16.39P

Angular acceleration of cylinder a = ${\alpha }_{\text{B}}\text{ = 9}{\text{.659 rad/s}}^2$

Angular acceleration of cylinder A = ${\text{α}}_A\text{ = 61}{\text{.824 rad/s}}^2$

Explanation of Solution

Given:

weight of Cylinder A, w_a = 5 lb radius of cylinder A = 4 in.

weight of cylinder B, w_b = 20 lb

radius of cylinder B, r_B = 8 in.

Magnitude of Force pulling the belt = 3.6 lb

Coefficient of static friction = 0.5

Coefficient of kinetic friction = 0.4

Concept used:

Condition of non-slipping,

Friction force due to static friction must be greater than the pulling force on each cylinder.

F_s> F_A, F_s>F_B

$\begin{array}{l}{\text{F}}_s{\text{ = μ}}_{\text{s}}{\text{N> F}}_{\text{A}}\\ {\text{F}}_s{\text{ = μ}}_{\text{s}}{\text{N> F}}_{\text{B}}\\ {\text{ In case of slipping, maximum tangential force between a cylinder and belt is F}}_s\text{ , }\\ {\text{otherwise it F}}_A{\text{ and F}}_B\text{, respectively for cylinder A and B}\end{array}$

Mass moment of inertia for a disk is given by-

$\text{I = }\frac{1}{2}{\text{mr}}^2,\text{ where r is the radius of disk }$

The tangential force acting on a cylinder will provide the angular acceleration to the cylinder. Therefore,

$\begin{array}{l}\text{Summation of moments applied on it = Mass Moment of inertia }\times \text{ angular acceleration}\\ \Sigma \text{M = Iα }\end{array}$

Calculation:

Mass of cylinder A = $\frac{\text{5lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.15528 lb-s}}^{\text{2}}\text{/ft}$

Mass of cylinder B = $\frac{\text{20lb}}{\text{32}{\text{.2ft/s}}^{\text{2}}}\text{ = 0}{\text{.62112 lb-s}}^{\text{2}}\text{/ft}$

Mass moment of inertia of cylinder A = $\frac{1}{2}{\text{m}}_{\text{A}}{r}_{\text{A}}^{\text{2}}\text{ = }\frac{1}{2}\text{(0}{\text{.15528 lb-s}}^{\text{2}}\text{/ft)×(}\frac{4}{\text{12}}{\text{)}}^{\text{2}}\text{ = 8}{\text{.6267×10}}^{\text{-3}}{\text{lb×s}}^{\text{2}}\text{ft}$

Mass moment of inertia of cylinder B = $\frac{1}{2}{\text{m}}_B{r}_B^{\text{2}}\text{ = }\frac{1}{2}\text{(0}{\text{.62112 lb-s}}^{\text{2}}{\text{/ft)×(8/12)}}^{\text{2}}\text{ = 0}{\text{.13803 lb×s}}^{\text{2}}\text{ft}$

Assuming slipping,

$\begin{array}{l}{\text{coefficient of kinetic friction is applicable μ}}_k\text{= 0}\text{.4}\\ {\text{F}}_s{\text{ = F}}_B{\text{= μ}}_k\text{N= 0}\text{.4}\times \text{5 lb = 2 lb}\end{array}$

For cylinder B,

$\begin{array}{l}{\text{ΣM}}_{\text{B}}{\text{ = I}}_{\text{B}}{\text{α}}_{\text{B}}\\ {\text{F}}_{\text{B}}{\text{×r}}_B{\text{= I}}_{\text{B}}{\text{α}}_{\text{B}}\\ 2\times (\frac{8}{12})\text{= (0}{\text{.13803) lb×ft×s}}^{\text{2}}{\text{ × (α}}_B\text{)}\\ {\alpha }_{\text{B}}\text{ = 9}{\text{.659 rad/s}}^2\end{array}$

Next, for the belt,

$\begin{array}{l}{\text{ΣF = P- F}}_{\text{A}}{\text{-F}}_B\text{ = 0}\\ \text{3}{\text{.6 lb - F}}_{\text{A}}\text{- 2lb = 0 }\\ {\text{ F}}_{\text{A}}\text{ = 1}\text{.6 lb}\end{array}$

Since ${\text{ F}}_{\text{A}}$ < ${\text{ F}}_s$

There is no slipping on cylinder A

$\begin{array}{l}{\text{ΣM}}_{\text{A}}{\text{ = I}}_{\text{a}}{\text{α}}_{\text{A}}\\ {\text{F}}_{\text{A}}{\text{×r}}_{\text{A}}{\text{= I}}_{\text{A}}{\text{α}}_{\text{A}}\\ \text{1}\text{.6lb ×(4/12 ft) = (8}{\text{.6267×10}}^{\text{-3}}{\text{) lb×ft×s}}^{\text{2}}{\text{ × (α}}_A\text{)}\\ {\text{α}}_A\text{ = 61}{\text{.824 rad/s}}^2\end{array}$

Conclusion:

Angular acceleration of cylinder a = ${\alpha }_{\text{B}}\text{ = 9}{\text{.659 rad/s}}^2$

Angular acceleration of cylinder A = ${\text{α}}_A\text{ = 61}{\text{.824 rad/s}}^2$