High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Grade 12 Physics Textbook > Dynamics

Inclined Surfaces

Normal ForcePrevious Section
Tension Next Section
Popular Courses
Find My Course
0:00 / 0:00

Inclined Surfaces


  • Sliding down an inclined plane is another example of motion with constant acceleration.
  • Source of the motion is still gravity of earth but only part of it is affecting the sliding of the object
  • To find the acceleration of an object over an inclined surface, one should first find the acceleration using the second law of Newton

Wize Tip
For inclined surfaces, it is easier to choose the coordinates such that x-axis is parallel to the surface and y-axis is perpendicular to the surface.


Example: Free-body diagram below is drawn for a skier of mass m gliding down a slope at angle θ. We have assumed the snow is frictionless.










Wize Tip
Since the sliding object remains in contact with the plane, there is no acceleration in the y-axis (if we define our axes along the plane).
PAGE BREAK

  • If the only force acting on the object is gravity, the acceleration of the sliding object along the direction of the inclined surface is:
ax=gsin⁡θa_x=g\sin\theta
  • where g=9.81ms2g=9.81\frac{m}{s^2} is the gravitational acceleration of earth
  • θ\theta is the angle between the inclined plane and the horizontal surface.
  • If friction is present, it always opposes movement (and therefore the force of gravity in this case)

0:00 / 0:00

Example: Finding Minimum Angle of Sliding


If the coefficient of static friction between the block and the ramp is 0.40.4, find the minimum angle at which the block will slide down the slope.




Solution:


First let's look at the free body diagram:
Equations of Motion

Need to decompose FgF_ginto the xxand yycoordinatesFx=Fgsin⁡θF_x=F_g\sin\theta and Fy=Fgcos⁡θF_y=F_g\cos\theta
∑Fx=0=−Ff+Fx∑Fy=0=N−Fy\begin{array}{l}\sum_{ }^{ }F_x=0=-F_f+F_x \\\sum F_y=0=N-F_y\end{array}

Since we are looking for the moment at which box starts to move, we are talking about threshold of motion for which static friction is maximum and equal to μsN\mu_s N

From the second equation above, N=Fy=mgcos⁡θN=F_y=mg\cos \theta

Substituting Ff=μsN=μsmgcos⁡θF_f=\mu_sN=\mu_s mg \cos\thetain the first equation above:
μsmgcos⁡θ=mgsin⁡θ→μs=tan⁡θθ=tan⁡−1(0.4)=21.8°\begin{array}{l}\mu_smg\cos\theta=mg\sin\theta\to\mu_s=\tan\theta\\\theta=\tan^{-1}(0.4)=\boxed{21.8\degree}\end{array}


Two blocks are connected via a pulley as shown below on a 45-degree inclined surface. The static and kinetic coefficient of friction between the m2m_2 and the surface are 0.45 and 0.25. The system is moving with constant speed such that m1m_1 is going down. What is m1/m2m_1/m_2?

PAGE BREAK