Force Vectors: Find Horizontal Acceleration & Direction for 5.00-kg Block

  • Thread starter Thread starter MakeItThrough
  • Start date Start date
  • Tags Tags
    Force Vectors
AI Thread Summary
Two forces, F1 and F2, act on a 5.00-kg block, with magnitudes of 30.0 N and 45.5 N, respectively. The horizontal component of F1 is calculated as 12.68 N, while F2 is 45.5 N, leading to a net force of 32.82 N in the horizontal direction. Using Newton's second law, the horizontal acceleration is determined to be 6.564 m/s². The direction of the acceleration is to the right, as F2 exceeds the horizontal component of F1. Understanding that acceleration follows the direction of the net force clarifies the problem.
MakeItThrough
Messages
13
Reaction score
0

Homework Statement


http://www.webassign.net/CJ/04_11.gif
Two forces, F1 and F2, act on the 5.00-kg block shown in the drawing. The magnitudes of the forces are F1 = 30.0 N and F2 = 45.5 N. What is the horizontal acceleration (magnitude and direction) of the block? Direction... Left or Right?

Homework Equations


F = mass x acceleration
acceleration = F / Mass

The Attempt at a Solution


I got two x and y components:
F1x = cos65 x 30 = 12.68 N
F1y = sin65 x 30 = 27.19 N
I do not know how to calculate the acceleration, please help and explain. Thanks in advance.

*I figured out the horizontal acceleration, but how do you know which direction it goes? Left or right?
 
Last edited:
Physics news on Phys.org
Theer are a couple of other forces in the y direction not shown, but that block isn't going anywhere in the y direction, so you need not concern yourself with y direction forces in this problem . You have F1x correct, what about F2x? Then look in the x direction using Newton's 2nd law to solve for the acceleration (magnitude AND direction) in the x direction.
 
ahh I figured it out. F1x is 12.68 N and F2 is 45.5 N. I just subtract 12.68 from 45.5 and that gives me 32.82 N. The mass of 5 kg is given.
Plugging this into the formula F=ma
32.82 / 5 kg = 6.564 m/s^2.
However, I'm not sure which direction the horizontal acceleration goes. Left or Right? How can you tell?
 
MakeItThrough said:
ahh I figured it out. F1x is 12.68 N and F2 is 45.5 N. I just subtract 12.68 from 45.5 and that gives me 32.82 N. The mass of 5 kg is given.
Plugging this into the formula F=ma
32.82 / 5 kg = 6.564 m/s^2.
However, I'm not sure which direction the horizontal acceleration goes. Left or Right? How can you tell?
The acceleration is always in the direction of the net force. Since F2 is greater than the horizontal component of F1, then the horizontal net force acts to the ____ and thus the acceleration is to the ____.
 
PhanthomJay said:
The acceleration is always in the direction of the net force. Since F2 is greater than the horizontal component of F1, then the horizontal net force acts to the ____ and thus the acceleration is to the ____.

Thank you! This is now starting to make sense!
(left)
 
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