Calculate acceleration of object with applied force and friction

AI Thread Summary
A worker pulls a crate with a force of 450 N at a 38-degree angle, while a frictional force of 125 N opposes the motion. The horizontal component of the pulling force is calculated as 354 N, and after accounting for friction, the effective force is 229 N. The acceleration of the crate, with a mass of 310 kg, is determined using the formula a = F/m, resulting in an acceleration of 0.74 m/s². The initial miscalculation arose from incorrectly considering the net force in both x and y directions instead of focusing solely on the horizontal component. The final correct acceleration reflects the influence of friction and the angle of the applied force.
username12345
Messages
47
Reaction score
0

Homework Statement



A worker drags a crate across a factoryt floor by pulling on a rope tied to the crate. The worker exerts a force of 450 N on the rope, which is inclined 38 degrees to the horizontal, and the floor exerts a horizontal frictional force on 125 N that opposes the motion. You can assume the crate doesn't leave the ground.

Calculate the acceleration of the crate (mass = 310 kg).

Homework Equations



F = ma

The Attempt at a Solution



Broke the force of the pull into x and y components:

Fx = 450 cos 38 = 354 N
Fy = 450 sin 38 = 227 N

Then the friction opposes motion, so subtract that from the x component

Fx = 354 - 125 = 229 N

so the net force is \sqrt{229^2 + 227^2} = 322 N

then, a = \frac{F}{m} = \frac{322}{310} = 1.0 ms^{-2} to 2 s.f.

However the answer given is 0.74 ms
 
Physics news on Phys.org
You're doing well until

so the net force is \sqrt{229^2 + 227^2} = 322 N

then, a = \frac{F}{m} = \frac{322}{310} = 1.0 ms^{-2} to 2 s.f.

However the answer given is 0.74 ms

The crate does not leave the ground so the only acceleration will be in the x-direction.
 
Ah, got it...

a = \frac{F_x}{m} = \frac{229}{310} = 0.74

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