Horizontal Force on a Ramp required to Accelerate a Mass

AI Thread Summary
To determine the horizontal force needed to accelerate a 5.57 kg box up a 38-degree ramp with a constant acceleration of 4.30 m/s², the calculations involve both gravitational and frictional forces. The initial calculation yielded a force of 23.951 N for acceleration, while the frictional force was calculated as 12.044 N. The total horizontal force required was then computed as 45.7 N, but there is uncertainty regarding this result. Clarifications on the calculations suggest that the horizontal component of the acceleration and the frictional force must be accurately combined to find the correct total force. The discussion emphasizes the importance of correctly applying the equations of motion and the effects of friction in inclined plane problems.
GiantCube
Messages
2
Reaction score
0

Homework Statement



A 5.57 kg box sits on a ramp that is inclined at 38.0^\circ above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.28.

What horizontal force is required to move the box up the incline with a constant acceleration of 4.30 m/s^2?

Homework Equations



Calculation F. F=ma is important. Trig functions. Calculation of F caused by friction.

The Attempt at a Solution



I calculated the force required to move the object up the ramp using a horizontal force; m*a/cos(38), which is 23.951N.

I calculated the force caused by the μk, which is sin 52*mg, which is 12.044 N

I added these, and found the horizontal component required, which is those added/ cos 28, which gives 45.7N

I am fairly certain I'm doing this wrong though, and I would appreciate any help.

Thanks!
 
Physics news on Phys.org
Welcome to PF, GiantCube.
I get the 12 and the 24 but not the 45.7. It would be nice to see your F = ma statement with the three things added to get the total force. We may have some differences in the signs.
 
Thanks. :)

The Force is the horizontal portion of the acceleration, so (m*a*cos(38)) must the force to get over the friction, so sin(52)*m*g. Which would be F= sin(52)*m*g + (u*m*a*cos(38))

... I'm fairly certain that's wrong, but not sure why...
 
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