1
$\begingroup$

Questions:

  1. For a block kept on a FRICTIONLESS FIXED INCLINE the acceleration down the plane is g*sin(A) (A is the angle of inclination) this acceleration can be further said to have components in the horizontal direction. But since gravity is only acting vertically downwards how can it cause an acceleration in the horizontal direction. Is the Normal from the surface of the incline the reason behind it. If so how?

  2. Is it okay to take components of a force, find the acceleration in the direction of that component and then further take components of the said acceleration.

$\endgroup$
  • $\begingroup$ Why would you think that you can't resolve the acceleration down a plane into x and y components? $\endgroup$ – Bob D Sep 28 at 17:02
  • $\begingroup$ @Bob D since acceleration down the plane is caused by the sin component of mg (in case of a fixed incline) taking a component of acceleration down the inclined plane in the horizontal direction leads me to believe that the force due to gravity is causing an acceleration in a direction perpendicular to it which is not possible. $\endgroup$ – Aditya Ahuja Sep 28 at 18:13
  • $\begingroup$ Would you then also say there is no vertical downward component of the acceleration down the plane? $\endgroup$ – Bob D Sep 28 at 18:57
  • $\begingroup$ And if there is, what would it be? $\endgroup$ – Bob D Sep 28 at 19:54
0
$\begingroup$

Gravity pushes the mass against the surface (which in this case cannot move). The surface pushes back with a normal force [mg cos(A)] which does have a horizontal component [mg cos(A) sin(A)] which causes the horizontal acceleration [ax = a cos(A)]. Knowing this, one can solve for a = g cos(A) sin(A) / cos(A), but you do not need a component of ax. In finding an acceleration or the component of an acceleration, be sure to consider all forces in that direction.

$\endgroup$
  • $\begingroup$ You should format your answer with MathJaX $\endgroup$ – Triatticus Sep 30 at 17:58
0
$\begingroup$

NET force is what is most important to body. Indeed in this case the object is moved down the surface by superposition of gravity and normal forces. Consider this picture : enter image description here

So net force acting on body is : $$ \vec{F}_{net} = \vec{F}_{w} + \vec{N} \\= m\vec{g} + \hat{\textbf{n}}\space F_{w}\cos(\alpha) \\=\hat{\textbf{u}}\space mg + \hat{\textbf{n}}\space mg\cos(\alpha) \\=mg(\hat{\textbf{u}} + \hat{\textbf{n}}\space \cos(\alpha)) $$

where $\hat{\textbf{u}}, \hat{\textbf{n}}$ are unit vectors of gravity and normal forces respectively

So when $\alpha=0$, then unit vectors cancels each other out completely, making net force zero,
and when $\alpha=\pi/2$, net force becomes $m\vec{g}$ - returning total control to gravity.

$\endgroup$
0
$\begingroup$

I have deleted my original answer, submitting the following instead in order to more directly answer your questions. In order to provide a more logical overall answer, I will answer your second question first.

2) Is it okay to take components of a force, find the acceleration in the direction of that component and then further take components of the said acceleration.

Yes.

Forces and accelerations are vectors. Any vector can be resolved into orthogonal components. It doesn’t matter what the origin of the vector is, that is, the vector being resolved into components may be a component of a different vector.

But since gravity is only acting vertically downwards how can it cause an acceleration in the horizontal direction. Is the Normal from the surface of the incline the reason behind it. If so how?

Gravity is not causing an acceleration of the block in the horizontal (x) direction. As you have correctly surmised, it is the normal reaction of the wedge that causes the horizontal acceleration. To be more specific, it is the horizontal component of the reaction force exerted by the wedge on the block that is causing the horizontal acceleration of the block.

Fig 1 below shows the block sliding down the wedge. To the right of it is a free body diagram of the block. Per Newtons's third law, the wedge exerts and equal and opposite reaction force to the normal force the block applies to the wedge. Since the wedge is fixed, the structure prevents it from moving so that the net force on the wedge is zero.

The normal reaction force applied to the block is resolved into a horizontal (x) component to the right and an upward vertical (y) component on the FBD. The vertical (y) upward component is opposed by the downward force of gravity. The horizontal (x) component of the force is unopposed. It is the net force acting to the right and is solely responsible for the horizontal acceleration of the block.

Fig 2 summarizes the forces acting on the block. It shows the horizontal force acting on the block which gives it its horizontal acceleration, and the net downward (y) component of the force on the block, which gives it its downward acceleration.

From the equations we can make the following observations:

  1. The maximum horizontal acceleration of the block occurs at an angle of 45 degrees and equals $g/2$.

  2. The horizontal acceleration is zero at 0 and 90 degrees.

  3. The maximum downward vertical acceleration of the block occurs at an angle of 90 degrees.

  4. The downward acceleration is zero at 0 degrees.

Hope this helps.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.