Question

A plank with a mass $M=6.00kg$rests on top of two identical, solid, cylindrical rollers that have $R=5.00cm$and $m=2.00kg$(Fig. P10.87). The plank is pulled by a constant horizontal force $\overrightarrow{F}$of magnitude localid="1663659037594" $6.00N$applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank. (a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank. (b) Find the acceleration of the rollers at this moment. (c) What friction forces are acting at this moment?

Short answer

The solution is

a) The acceleration of the plank$0.800m/s^2$

b) The acceleration of the rollers$0.400m/s^2$

c ) The frictional forces$f_b=-0.200\text{\hspace{0.17em}N}$

Step-by-step solution

About the problem

Conceive: Consider the frictional force exerted by each roller and the rolling resistance exerted backward by the ground on each roller to conceptualize the situation.

CATEGORIZE: This issue falls under Newton's second law of motion.

Given information

We are given that

Mass of the plank$M=6.00\text{kg}$

Radius of the solid cylinders$R=5.00\text{cm}$

Mass of the cylindrical rollers$m=2.00\text{kg}$

Force with which the plank is pulled$F=6.00\text{\hspace{0.17em}N}$

Frictional force

The frictional force exerted by each roller is denoted by $f_t$. Let $f_b$be the backward rolling resistance of the ground on each roller.

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If the rollers are evenly spaced from the plank's ends, we get by applying Newton's second law of motion to the plank.

$\begin{array}{c}\Sigma F_x=m{a}_x\\ 6.00\text{N}-2f_t=(6.00\text{kg})a_p\end{array}$

Each roller's center $\left(\frac{a_p}{2}\right)$only slides forward half as far as the plank

$\left(\frac{\frac{a_p}{2}}{5.00cm}\right)=\left(\frac{a_p}{0.100m}\right)$

The initial acceleration of the plank

1. We get for each cylinder.

$\begin{array}{c}\Sigma F_x=m{a}_x\\ +f_t-f_b=(2.00\text{kg})\left(\frac{a_p}{2}\right)\end{array}$

The net torque is calculated using the formula

role="math" localid="1663658837253" $\begin{array}{c}\Sigma \tau =I\alpha \\ f_t(5.00\text{cm})+f_b\left(5.00\text{\hspace{0.17em}cm}\right)=\frac{1}{2}(2.00\text{kg}){\left(5.00\text{\hspace{0.17em}cm}\right)}^2\left(\frac{a_p}{10.0\text{cm}}\right)\end{array}$

We get,

$f_t+f_b=\left(\frac{1}{2}\text{kg}\right)a_p$

When we combine the two equations above, we obtain

$2f_t=(1.50\text{kg})a_p$

Calculating the acceleration,

$\begin{array}{c}(6.00\text{N})-(1.50\text{kg})a_p=(6.00\text{kg})a_p\\ a_p=\left(\frac{6.00\text{N}}{7.50\text{kg}}\right)\\ =0.800{\text{m/s}}^{\text{2}}\end{array}$

The acceleration of the rollers

b) The roller's acceleration will be

$\begin{array}{l}a=\frac{a_p}{2}\\ =0.400{\text{m/s}}^{\text{2}}\end{array}$

Friction forces acting at this moment

c) Consider the free body diagram in this section.

The frictional forces were calculated as

$\begin{array}{l}2f_t=(1.50\text{kg})a_p\\ 2f_t=(1.50\text{kg})(0.800{\text{m/s}}^2)\\ f_t=0.600\text{N}\end{array}$

On the other hand

$\begin{array}{c}0.600\text{\hspace{0.17em}N}+f_b=(\frac{1}{2}\text{kg})(0.800{\text{m/s}}^{\text{2}})\\ f_b=-0.200\text{N}\end{array}$

The negative sign implies that the horizontal ground force on each roller is 0.200 forward, not backward, as we previously anticipated.