Question

A loaded penguin sled weighing ${}^{\mathbf{80}\mathbf{}\mathbf{N}}$rests on a plane inclined at angle ${}^{\mathbf{\theta }\mathbf{=}{\mathbf{20}}^{\mathbf{0}}}$to the horizontal (Fig. 6-23). Between the sled and the plane, the coefficient of static friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{25}}$, and the coefficient of kinetic friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{15}}$. (a) What is the least magnitude of the force parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude ${}^{\mathbf{F}}$that will start the sled moving up the plane? (c) What value of ${}^{\mathbf{F}}$is required to move the sled up the plane at constant velocity?

Short answer

(a) The least magnitude of force ${}^F$that prevents the sled moving down is ${}^{8.6\mathrm{N}}$

(b) The minimum magnitude of F that will start the sled moving up the plane is ${}^{46\mathrm{N}}$

(c) The value of F, that is required to move the sled up the plane at constant velocity is ${}^{39\mathrm{N}}$

Step-by-step solution

Given

Weight,${}^{\mathrm{W}=80\mathrm{N}}$

Coefficient of static friction,${}^{{\mathrm{\mu }}_{\mathrm{s}=0.25}}$

Coefficient of kinetic friction,${}^{{\mathrm{\mu }}_{\mathrm{k}}=0.15}$

Inclined angle,${}^{\mathrm{\theta }={20}^0}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

$F_{net}=\sum ma$

where,Fis the net force,mis mass andais an acceleration.

Determining the free body diagram

Free body Diagram of sled over the inclined plane:

(a) Determining the least magnitude of the force parallel to the plane

First, from the weight, calculate the mass of the system,

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{80\mathrm{N}}{9.81} \\ =8.2\mathrm{kg} \end{gathered}$$

By using Newton’s 2nd law along vertical direction (along y),

$\sum F_y=m{a}_y$

since, block is not moving upward, $a_y=0$

$$\begin{gathered} N-F_g\cos \left(\theta \right)=0 \\ N=F_g\cos \left(\theta \right) \end{gathered}$$

Relation between static frictional force and normal force is,

$f_s={\mu }_{s}N=N_s{F}_g\cos \left(\theta \right)$ (i)

When sled is moving down frictional force is upward,

Now, applying Newton’s 2nd along the x direction, for fig. a,

$\sum F_s=m{a}_x$

since, block is not moving, localid="1654149316961" $a_x=0$

$$\begin{gathered} F+f_s-F_g\sin \left(\theta \right)=0 \\ F=f_g\sin \left(\theta \right)-f_s \end{gathered}$$

By using equation (i) in above equation,

$$\begin{gathered} F=F_g\sin \left(\theta \right)-{\mu }_s{F}_g\cos \left(\theta \right) \\ F=F_g(\sin \left(\theta \right)-{\mu }_s\cos \left(\theta \right)) \\ =mg(\sin \left(20\right)-(0.25\left)\cos \left(20\right)\right) \\ =8.6N \end{gathered}$$

Thus, the least magnitude of force ${}^F$that prevents the sled moving down is ${}^{8.6\mathrm{N}}$

(b) Determining the minimum magnitude of F that will start the sled moving up the plane

To find the F to start the sled moving up is, again we have to use Newton’s 2nd law along y and x direction, but in this case as the sled moving up frictional force is downward.So, refer to fig. b.

Along vertical direction (y),

$\sum F_y=m{a}_y$

since block is not moving upward,$a_y=0$

$$\begin{gathered} N-F_g\cos \left(\theta \right)=0 \\ N=F_g\cos \left(\theta \right) \end{gathered}$$

Then static frictional force,

$f_s={\mu }_{s}N={\mu }_s{F}_g\cos \left(\theta \right)$

Along horizontal direction (x),

$\sum F_x=m{a}_x$

since, block is not moving, $a_x=0$

$$\begin{gathered} F-f_s-F_g\sin \left(\theta \right)=0 \\ F=F_g\sin \left(\theta \right)+f_s \\ =F_g(\sin \theta +{\mu }_s\cos \theta ) \\ =46N \end{gathered}$$

Hence, the minimum magnitude of F that will start the sled moving up the plane is ${}^{46\mathrm{N}}$.

(c) Determining thevalue of F, that is required to move the sled up the plane at constant velocity

To find ${}^F$to move the sled up the plane at constant velocity, use kinetic frictional force,

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_k{F}_g\cos \left(\theta \right) \end{gathered}$$

Along horizontal direction (x),

$\sum F_x=m{a}_x$

since, block is moving with constant velocity, $a_x=0$

$$\begin{gathered} F-f_k-F_g\sin \left(\theta \right)=0 \\ F=F_g\sin \left(\theta \right)+f_k \\ =F_g(\sin \theta +{\mu }_k\cos \theta ) \\ =39N \end{gathered}$$

Hence, the value of F, that is required to move the sled up the plane at constant velocity is${}^{39\mathrm{N}}$