Question

Show that the acceleration of any object down a frictionless incline that makes an angle $\theta$ with the horizontal is $a = g \sin{\theta}$ . (Note that this acceleration is independent of mass.)

Final Answer

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### Solution video

# OpenStax College Physics for AP® Courses, Chapter 5, Problem 8 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We are going to show that an object sliding down a frictionless incline at an angle of

*Θ*is going to have an acceleration equal to acceleration due to gravity multiplied by*sin Θ*. So we have to draw a free-body diagram and there's going to be only... well, I guess there's the normal force here as well but it's only the gravity that's important because this normal force does not get translated into any friction here as would normally be the case because we assume that the incline is frictionless. Okay! So we have gravity straight down— I defined the coordinate system so that down the ramp is positive— and we have a y-component of gravity and we have a x-component of gravity or a perpendicular component and a parallel component and it's this x-component that we are concerned with. And this angle in here I have labeled it as*Θ*but let's pretend that we don't really know what the angle is: let's call it*β*and I'll show you why it equals this incline angle of*Θ*. So we know this is a right triangle here when we consider this triangle like here and with this being a right triangle, we'll call this angle here*α*and we'll figure out what*α*is when we know that these three angles have to add up to 180 so that means*Θ*plus*α*plus 90 equals 180 in which case*α*is 180 minus this 90 which is 90 and then minus this*Θ*as well so that's*α*. And then we know that*α*plus*β*has to work out to 90 because this dotted line is perpendicular to the incline so let's write that down:*α*plus this unknown angle*β*equals 90 and then we'll solve for*β*by subtracting*α*from both sides and we have*β*then is 90 minus*α*. And now let's substitute in what*α*is from up here: it is 90 minus*Θ*. So we have 90 minus 90 minus*Θ*, the*Θ*becomes positive, the 90's cancel and we are left with*Θ*. So this angle here is*Θ*so we don't have to call it*β*anymore, we can call it*Θ*now. Okay! So the x-component of this gravity is going to equal the mass times acceleration because there are no other forces acting horizontally and this x-component of gravity then is the gravity*F g*multiplied by the*sin*of this angle*Θ*because we are trying to find the opposite leg of this right triangle and we use*sin*of the angle multiplied by the hypotenuse to get it and this gravity is mass times acceleration due to gravity,*mg*. So we can replace*F g x*with*mgsin Θ*equals to*ma*and then divide both sides by*m*and switch the sides around and we get the expression we wanted that acceleration equals*g*times*sin Θ*.