Question

Suppose you have a 120-kg wooden crate resting on a wood floor. (a) what maximum force can you exert horizontally on the crate without moving it? (b) if you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be?

Answer 1

According the attached picture so we can assume that :

The coefficient of friction wood on wood are:

static friction μs = 0.5

dynamic friction μk = 0.2

a)The answer is:

F= 588.6 N

The explanation:

The maximum force  you can exert horizontally on the crate without moving it is equal:

Fsmax = μs * M * g

when μs is the static friction

and M i she mass

g is the gravitational force by substituted:

Fsmax = 0.5 * 120Kg * 9.81 m/s2

           = 588.6 N

when F= Fsmax

So the answer is F = 588.6 N

b)The answer is:

a= 2.943 m/s2

The explanation:

Fsmax is the force pushing the block , and there is a force of friction trying to stop the block .

So Fk = μk * N

when Fk = force of kinetic friction

μk = coefficient of kinetic friction of wood on wood

N= Normal force by substitution:

Fk= 0.2 * 120k *9.81 m/s2 = 235.44 N

so F = Fsmax - Fk

        588.6 - 235.44 = 353.16 N

when F = m*a

so 353.16 = 120Kg * a

so a ( the acceleration) = 353.16/120kg

                                      = 2.943 m/s2

Answer 2

(a)The maximum horizontal force that can be exerted on the crate without moving it is $\boxed{588{\text{ N}}}$.

(b) The acceleration of the crate due to this maximum force is $\boxed{2.94{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$ or $\boxed{2.9{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$.

Further explanation:

The friction force acting on the wooden crate is acting in the direction opposite to the motion of the wooden crate and therefore, the friction force opposes the relative motion of the wooden crate on the wooden floor.

Given:

The mass$\left( m \right)$ of the wooden crate is $120{\text{ kg}}$.

The static coefficient of friction $\left( {{\mu _s}} \right)$ is $0.5$.  

The dynamic coefficient of friction $\left( {{\mu _k}} \right)$ is $0.2$.  

Formula and concept used:

Weight of the crate can be calculated as,

$W=mg$                                               ........ (1)

Here, $g$ is the acceleration due to gravity.

The maximum force is applied on the crate but the crate does not move from its position, if we further increase the force on the crate, the crate will start to move.

The motion of the case is opposed by the friction force acting opposite to the direction of applied force. Therefore, this is the case of limiting friction.

The maximum force on the crate can be calculated as,

$\boxed{{F_{\max }}={\mu _s}W}$                                                   …… (2)

Force required to keep wooden crate in continuous motion is,

$\boxed{F={\mu _k}W}$                                                      …… (3)

But, during the motion force applied on the crate is the maximum. The excess force applied on the crate results in the motion of the crate.

The excess force applied on the crate which makes the crate to accelerate is:

By using Newton’s law of motion,

$ma={F_{\max }} - F$              

Rearrange the above expression.          

$\boxed{a=\frac{{{F_{\max }} - F}}{m}}$                                                             …… (4)

Calculation:

(a).

Substitute the value of $m$ as $120{\text{ kg}}$ and $g$ as $9.8\text{ m}/\text{s}^2$ in equation (1).

$\begin{aligned}\\W&=120 \times 9.8 \hfill \\&=1176{\text{ N}} \hfill \\ \end{aligned}$

Substitute the value of $W$ as $1176{\text{ N}$ and ${\mu _s}$ as $0.5$ in equation (2)

$\begin{aligned}{F_{\max }}&=0.5 \times 1176 \hfill \\&=588{\text{ N}} \hfill \\ \end{aligned}$  

(b).

Substitute the value of ${\mu _k}$ as $0.2$ and value of $W$ as $1176{\text{ N}}$ in equation (3).

$\begin{aligned}\\F&=0.2 \times 1176 \hfill \\&=235.2{\text{ N}} \hfill \\ \end{aligned}$  

To determine the acceleration substitute the value of ${F_{\max }}$ as $588{\text{ N}}$ and the value of $F$ as $235.2{\text{ N}}$ in equation (4).

$\begin{aligned}\\a&=\frac{{588 - 235.2}}{{120}} \\&=\frac{{352.8}}{{120}} \\&=2.94{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}} \\ \end{aligned}$

Thus,

a)The maximum horizontal force that can be exerted on the crate without moving it is $\boxed{588{\text{ N}}}$.

(b) The acceleration of the crate due to this maximum force is $\boxed{2.94{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$ or $\boxed{2.9{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}$.

Learn more:

1. What is threshold frequency of cesium: brainly.com/question/6953278

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

Answer detail:

Grade: Senior school

Subject: Physics

Chapter: Friction

Keywords:

Wooden crate, maximum horizontal force, acceleration of the crate, friction, dynamic friction, continuous motion, limiting friction, weight, maximum force, 2.9 m/s2, 2.94m/s2, 2.9 m/s^2, 2.94m/s^2, 588N, 588N, 588Newton.