Inclined Plane Problem Two Parts

AI Thread Summary
The discussion revolves around solving an inclined plane physics problem involving a block sliding down a frictionless surface at an angle of 24.5°. The initial calculation for acceleration was incorrectly derived, leading to confusion about the application of Newton's second law. The correct approach involves resolving the gravitational force into components parallel and perpendicular to the incline. The final speed of the block at the bottom of the incline was calculated using the incorrect acceleration, prompting a need to reassess the force components. Understanding how to properly calculate these components is crucial for solving the problem accurately.
sugarntwiligh
Messages
24
Reaction score
0

Homework Statement


The block shown in Fig. 4-48 lies on a smooth plane tilted at an angle θ = 24.5° to the horizontal. Ignore friction.

http://www.webassign.net/giancoli5/4_48.gif (visit this site for picture)

(a) Determine the acceleration of the block as it slides down the plane.

(b) If the block starts from rest 9.50 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?


The Attempt at a Solution



a.)
9.8/sin(24.5)=23.6 m/s^2 which is the acceleration down the slope where the x=ma and the y=mg. Is this correct? It feels like I am forgetting something...
b.)
From the equation v^2=vi^2+2ad,
v^2=0+2*23.6*9.5
v=21.2m/s at the bottom of the incline.
 
Physics news on Phys.org
sugarntwiligh said:
a.)
9.8/sin(24.5)=23.6 m/s^2 which is the acceleration down the slope where the x=ma and the y=mg. Is this correct?
No, it's not. What forces act in the x direction? Apply Newton's 2nd law. (Sanity check: Compare your calculated acceleration with g. Does your answer make sense?)
b.)
From the equation v^2=vi^2+2ad,
v^2=0+2*23.6*9.5
v=21.2m/s at the bottom of the incline.
If you had the correct acceleration, that approach would work.
 
Doc Al said:
No, it's not. What forces act in the x direction? Apply Newton's 2nd law. (Sanity check: Compare your calculated acceleration with g. Does your answer make sense?)

If you had the correct acceleration, that approach would work.

I didn't use any value of x in my equation. But I am still confused about why its wrong, because sin(24.5)=(9.8m)/(ma). Where mg is the opposite and ma is the hypotenuse. And then solving for a gives me the same answer.

My answer would make more sense if I did 9.8sin(24.5)=4.06, and I think this is the right way to go, but why?
 
sugarntwiligh said:
I didn't use any value of x in my equation. But I am still confused about why its wrong, because sin(24.5)=(9.8m)/(ma). Where mg is the opposite and ma is the hypotenuse. And then solving for a gives me the same answer.
To use a triangle to find force components, you must realize that the total force is "mg", so mg must be the hypotenuse of your right triangle. (That right triangle is not simply the incline.)

My answer would make more sense if I did 9.8sin(24.5)=4.06, and I think this is the right way to go, but why?
You need to learn how to find the components of the weight parallel and perpendicular to the incline surface. Read this: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e.html"
 
Last edited by a moderator:
Ohhhhhhhhh I get it! Thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top