A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.
What is the magnitude of the frictional force exerted on the mug?
What is the minimum coefficient of static friction required to keep the mug from sliding?
Answers
The minimum coefficient of static friction required to keep the mug from sliding is 0.31
Further explanationAcceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
This problem is about Newton's Law of Motion.
Given:
θ = 17°
Unknown:
μ = ?
Solution:
[tex]\Sigma Fy = ma[/tex]
[tex]N - w \cos \theta = m (0)[/tex]
[tex]N - w \cos \theta = 0[/tex]
[tex]\boxed {N = w \cos \theta}[/tex] → Equation 1
[tex]\Sigma Fx = ma[/tex]
[tex]w \sin \theta - f = m (0)[/tex]
[tex]w \sin \theta - f = 0[/tex]
[tex]f = w \sin \theta[/tex]
[tex]\mu N = w \sin \theta[/tex]
[tex]\mu w \cos \theta = w \sin \theta[/tex] ← Equation 1
[tex]\mu = (w \sin \theta) \div ( w \cos \theta )[/tex]
[tex]\mu = \tan \theta[/tex]
[tex]\mu = \tan 17^o[/tex]
[tex]\mu \approx 0.31[/tex]
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Subject: Physics
Chapter: Dynamics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Mug , Friction , Coefficient , Static
The magnitude of frictional force exerted on the mug is [tex]\fbox{\begin\\({2.865}\right))m\end{minispace}}[/tex] and the coefficient of static friction is [tex]\fbox{0.31}[/tex].
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 [tex]{17^ \circ }[/tex].
The acceleration due to gravity is [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex].
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.
[tex]\fbox{\begin\\{F_{{\text{net}}}} = mg\sin \theta\end{minispace}}[/tex] …… (1)
Here,[tex]m[/tex] is the mass of the mug, [tex]g[/tex] is the acceleration due to gravity and[tex]\theta[/tex] 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.
[tex]{F_s} = {\mu _s}N[/tex] ..…. (2)
Here,[tex]{\mu _s}[/tex] is the coefficient of static frictional force and [tex]N[/tex] 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.
[tex]N-mg\cos\theta=0[/tex]
Rearrange the above expression.
[tex]N=mg\cos\theta[/tex]
Substitute[tex]mg\cos\theta[/tex] for [tex]N[/tex] in equation (2).
[tex]{F_s} = {\mu _s}\left( {mg\cos \theta } \right)[/tex]
Here static frictional force is same as the frictional force exerted on the mug.
Substitute [tex]mg\sin\theta[/tex] for[tex]{F_s}[/tex] in above expression and rearrange it.
[tex]{\mu _s} = \dfrac{{\left( {mg\sin \theta } \right)}}{{\left( {mg\cos \theta } \right)}}[/tex]
Rearrange the above expression.
[tex]\fbox{\begin\\{{\mu _s} = \tan \theta}\end{minispace}}[/tex] …… (3)
Substitute[tex]{17^ \circ }[/tex]for [tex]\theta[/tex] and [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex]in equation (1).
[tex]\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}[/tex]
Substitute [tex]{17^ \circ }[/tex]for [tex]\theta[/tex] in equation (3).
[tex]\begin{gathered}{\mu_s}=\tan\left( {{{17}^\circ }}\right)\\= 0.306 \\\end{gathered}[/tex]
Thus, the coefficient of static friction is [tex]0.31[/tex] and the magnitude of frictional force is [tex](2.865)m[/tex].
Learn more:
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Friction, acceleration, carpeted floor, mug, inclined plane, force, relative motion, motion, normal reaction, net force, mass, surface, oppose, 9.8 m/s2, 17 degree, 0.67 m/s^2, 0.3006, 0.31, (2.865)m, (2.87)m.
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