Question

A 100 Nforce, directed at an angle$\mathit{\theta }$above a horizontal floor, is applied to a 25.0 kgchair sitting on the floor. If$\mathit{\theta }\mathbf{=}\mathbf{0}\mathbf{°}$, what are (a) the horizontal component${\mathit{F}}_{\mathbf{h}}$of the applied force and (b) the magnitudeNof the normal force of the floor on the chair? If$\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$, what are (c)${\mathit{F}}_{\mathbf{h}}$and (d)${\mathit{F}}_{\mathbf{N}}$? If$\mathit{\theta }\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$, what are (e)${\mathit{F}}_{\mathbf{h}}$and (f)${\mathit{F}}_{\mathbf{N}}$? Now assume that the coefficient of static friction betweenchair and floor is 0.420. Does the chair slide or remain at rest if$\mathit{\theta }$is (g)$\mathbf{0}\mathbf{°}$, (h)$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$, and (i)$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$?

Short answer

1. The horizontal component of applied force is 100 N.
2. The magnitude of normal force of the floor on the chair is 245 N.
3. $F_h\mathrm{for}\mathrm{\theta }=30.0°\mathrm{is}86.60\mathrm{N}.$
4. $F_N\mathrm{for}\mathrm{\theta }=30.0°\mathrm{is}195\mathrm{N}.$
5. $F_h\mathrm{for}\mathrm{\theta }=60.0°\mathrm{is}50\mathrm{N}.$
6. $F_N\mathrm{for}\mathrm{\theta }=60.0°\mathrm{is}158.4\mathrm{N}.$
7. The crate remains at rest, if $\mathrm{\theta }\mathrm{is}0°$.
8. The crate slides, if $\mathrm{\theta }\mathrm{is}30°$.
9. The crate must remain at rest, if $\mathrm{\theta }\mathrm{is}60°$.

Step-by-step solution

Given data

The force is, F = 100 N.

The mass is,m = 25 kg

Understanding the concept

The frictional force is given by the product of the coefficient of friction and the normal reaction.

Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Using these concepts, the problem can be solved.

(a) Calculate the horizontal component Fh of the applied force

The angle between the exerted force and horizontal is zero, $\theta =0°$.

The horizontal component will be in the x direction, and it can be written as,

$$\begin{gathered} F_h=F_x \\ {F}_h=F\cos \theta \left(1\right) \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F_h=100\mathrm{N}\cos 0° \\ =100\mathrm{N} \end{gathered}$$

Thus, the horizontal component of applied force is 100 N.

(b) Calculate the magnitude FN of the normal force of the floor on the chair

Since there is no vertical acceleration, the application of Newton’s second law in the y direction as,

$$\begin{gathered} F_N+F_y=mg \\ {F}_N+F\sin \theta =mg \\ {F}_N=mg-F\sin \theta \left(2\right) \end{gathered}$$

Substituting values in the above expression, and we get,

$$\begin{gathered} F_N=25\mathrm{kg}\times \times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =245\mathrm{N} \end{gathered}$$

Thus, the magnitude of normal force of the floor on the chair is 245 N.

(c) Calculate Fh if θ=30.0° if

Now, for $\theta =30°$.

Substituting values in the equation (1), and we get,

$$\begin{gathered} F_h=100\cos 30° \\ =100\times 0.866 \\ =86.60\mathrm{N} \end{gathered}$$

Thus, $F_h\mathrm{for}\mathrm{\theta }=30.0°\mathrm{is}86.60\mathrm{N}$.

(d) Calculate FN if θ=30.0° 

Substituting values in equation (2), and we get,

$$\begin{gathered} F_N=25\times 9.8-100\sin 30° \\ =245-100\times .5 \\ =195\mathrm{N} \end{gathered}$$

Thus, localid="1661152253657" $F_N\mathrm{for}\mathrm{\theta }=30.0°\mathrm{is}195\mathrm{N}$.

(e) Calculate Fh if θ=60.0° 

Now, for $\theta ={60}^{°}$.

Substituting values in the equation (1), and we get,

$$\begin{gathered} F_h=100\cos 60° \\ =100\times 0.5 \\ =50\mathrm{N} \end{gathered}$$

Thus, $F_h\mathrm{for}\mathrm{\theta }=60°\mathrm{is}50\mathrm{N}$.

(f) Calculate FN if θ=60.0°  if

Substituting values in equation (2), and we get,

$$\begin{gathered} F_N=25\times 9.8-100\sin 60° \\ =245-100\times 0.8660 \\ =158.4\mathrm{N} \end{gathered}$$

Thus, $F_N\mathrm{for}\mathrm{\theta }=60°\mathrm{is}158.4\mathrm{N}$.

(g) Figure out if the chair slide or remain at rest if θ is 0° is

The condition for the chair to slide can be written as,

$F_x>f_{s,max}$ (3)

And we know that,

$f_{s,max}={\mu }_s{F}_N$ (4)

Where coefficient of static friction is, ${\mu }_s=0.42,\mathrm{and}{f}_{s,max}$is the maximum frictional force.

For $\theta =0°$, from equation 4, we can write the expression as,

$$\begin{gathered} f_{s,max}=0.42\times 245\mathrm{N} \\ =102.9\mathrm{N} \end{gathered}$$

From subpart a,

$F_x=100\mathrm{N}$

$F_x<f_{s,max}$in this case. So the crate remains at rest.

Thus, the crate remains at rest, if $\theta$is $0°$.

(h) Figure out if the chair slide or remain at rest if θ is 30° is

For $\theta =30°$, from equation 4, we can write the expression as,

$$\begin{gathered} f_{s,max}=0.42\times 195\mathrm{N} \\ =81.9\mathrm{N} \end{gathered}$$

From subpart c,

$F_x=86.60\mathrm{N}$

$F_x>f_{s,max}$in this case. So the crate slides.

Thus, The crate slides; if $\theta =31°$.

(i) Figure out if the chair slide or remain at rest if θ is 60° 

For $\theta =60°$, from equation 4, we can write the expression as,

$$\begin{gathered} f_{s,max}=0.42\times 158.4\mathrm{N} \\ =66.52\mathrm{N} \end{gathered}$$

From subpart e,

$F_x=50\mathrm{N}$

$F_x<f_{s,max}$in this case. which means the crate must remain at rest.

Thus, the crate must remain at rest, if $\theta \mathrm{is}60°$.