Question

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Answer 1

Refer to the figure shown below.

g =  9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
 the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

Answer 2

The magnitude of frictional force exerted on the mug is $\fbox{\begin\\\left( {2.865} \right)m\end{minispace}}$ and the coefficient of static friction is $\fbox{\begin\\0.31\end{minispace}}$.

Further explanation:

The opposition that every object feels when it moves is known as the frictional force. This is an opposing force. It always acts in the opposite direction of the motion of the object.

Given:

The angle of inclination of plane is ${17^ \circ }$.  

The acceleration due to gravity is $9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}$.

Concept used:

The frictional force is defined as the opposition of motion of any object or body.  It is also defined as the product of coefficient of friction to the normal force exerted on the body.

The expression for the net force exerted on the mug is given as.

${F_{{\text{net}}}} = mg\sin \theta$                                                                         …… (1)

Here, m is the mass of the mug, g is the acceleration due to gravity and $\theta$  is the inclination angle.

The force that keeps the body at rest is known as the static friction force. It also acts in the opposite direction of motion of body.

The expression for the static frictional force is given as.

${F_s} = {\mu _s}N$                                                              ..…. (2)

Here, ${\mu _s}$ is the coefficient of static frictional force and N is the normal force exerted on the body.

When mug is placed on the inclined plane the static frictional force is equal to the frictional force. The vertical component of weight is balanced by the normal reaction of the mug.

The expression for the normal force is given as.

$N - mg\cos \theta=0$

 

Rearrange the above expression.

$N = mg\cos \theta$

 

Substitute $mg\cos \theta$  for N in equation (2).

${F_s}={\mu _s}\left( {mg\cos \theta } \right)$

 

Here static frictional force is same as the frictional force exerted on the mug.

Substitute $mg\sin \theta$ for ${F_s}$ in above expression and rearrange it.

${\mu _s} = \dfrac{{\left( {mg\sin \theta } \right)}}{{\left( {mg\cos \theta } \right)}}$

 

Rearrange the above expression.

${\mu _s} = \tan \theta$                                   …… (3)

Substitute ${17^ \circ }$ for$\theta$ and $9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for g in equation (1).

$\begin{aligned}{F_{{\text{net}}}}&=m\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\sin \left({{{17}^\circ}}\right)\\&=\left( {2.865}\right)m\\\end{aligned}$

 

Substitute ${17^ \circ }$ for $\theta$  in equation (3).

$\begin{gathered}{\mu _s}=\tan \left( {{{17}^ \circ }} \right)\\=30.06\\\end{gathered}$

 

Learn more:

1.  Motion under friction brainly.com/question/7031524.

2.Conservation of momentum brainly.com/question/9484203.

3. Net force on the body brainly.com/question/6125929.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Friction, acceleration, carpeted floor, mug, inclined plane, force, relative motion, motion, net force, oppose, 9.8 m/s2, 17 degree, 0.67 m/s^2, 30.056, 30.06, 31,(2.865)m,(2.87)m.