Solving for g on an inclined plane

AI Thread Summary
To solve for g on an inclined plane in terms of d, t, and theta, the equation derived is g = 2d/(sin theta * t^2). The acceleration a is expressed as g*sin theta, and with initial velocity Vi set to 0, the equation for distance d becomes d = 0.5(g*sin theta)t^2. The solution has been confirmed as correct by a participant in the discussion. This approach effectively relates gravitational acceleration to the parameters of the inclined plane scenario.
mysticbms
Messages
8
Reaction score
0

Homework Statement


Solve for g in terms of d, t, and theta on an inclined plane.



Homework Equations


Fnet=ma and d=Vit + .5a*t^2


The Attempt at a Solution


Solving for a on an inclined plane I get g*sin theta.

Using the equation for constant acceleration d=Vit + .5a*t^2 I solve for g.

Vi is 0 on an initial plane so I have d=.5(g*sin theta)t^2

g=2d/(sin theta*t^2)

Is this correct?
 
Physics news on Phys.org
mysticbms said:

Homework Statement


Solve for g in terms of d, t, and theta on an inclined plane.



Homework Equations


Fnet=ma and d=Vit + .5a*t^2


The Attempt at a Solution


Solving for a on an inclined plane I get g*sin theta.

Using the equation for constant acceleration d=Vit + .5a*t^2 I solve for g.

Vi is 0 on an initial plane so I have d=.5(g*sin theta)t^2

g=2d/(sin theta*t^2)

Is this correct?

Yes, correct.

ehild
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Rope between inclines with maximum length of hanging middle portion
Replies
46
Views
2K
Rolling without slipping down an inclined plane
Replies
18
Views
5K
Projectile motion on an inclined plane
Replies
7
Views
2K
Block on top of an inclined plane, with a rough surface, that moves with constant acceleration
Replies
41
Views
3K
Problem with Dynamic blocks on a fixed inclined plane
Replies
10
Views
2K
Acceleration of stacked blocks on an inclined plane
Replies
14
Views
2K
Kinematics on a frictionless incline plane
Replies
3
Views
2K
Two masses, two pulleys and an inclined plane
2
Replies
68
Views
12K
8.01 MIT OCW PS1.4: Throw and Catch (Kinematics)
Replies
4
Views
1K
Problem regarding projectile motion on a inclined plane
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top