Finding acceleration from two forces on a 14.1 kg block

  • Thread starter Thread starter xelda
  • Start date Start date
  • Tags Tags
    Force Vectors
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 9K views
xelda
Messages
23
Reaction score
0
I've been given this problem: Two forces act on a block of mass 14.1 kg. The first force has magnitude 28.5 N and is directed along the x axis. The second force has magnitude 6.7 N and is directed at an angle of 38.7 degrees to the x axis. Find (a) the magnitude and (b) the direction (i.e., the angle above the x-axis) of the block's acceleration. The block ONLY has the two indicated forces acting on it. Gravity DOES NOT play a role.

I keep doing this problem over, including saving the rounding until the end, but I still seem to be off by a little bit. If any of you could help me spot the error, I would appreciate it.

components of FA:
FAX[/size] = (28.5 N)(cos 38.7) = 22.24 N
FAY[/size] = (28.5 N)(sin 38.7) = 17.82 N

components of FB[/size]:
FBX[/size] = (6.7 N)(cos 38.7) = 5.23 N
F[size=-2]BY[/size] = (6.7 N)(sin 38.7) = 4.19 N

F[size=-1]RX[/size] = 22.24 + 5.23 = 27.47 N
F[size=-1]RY[/size] = 17.82 + 4.19 = 22.01 N

then I took the square root of the sum of FRX[/size] squared plus FRY[/size] squared and got 35.2 N.

I then divided this by 14.1 kg to find the acceleration and ended up with 2.496 m/s^2 but the program says it's incorrect. Close, but wrong. Any takers?
 

Discussion

Physics news on Phys.org
xelda said:
components of FA:
FAX[/size] = (28.5 N)(cos 38.7) = 22.24 N
FAY[/size] = (28.5 N)(sin 38.7) = 17.82 N
The 28.5 N force is directed along the x-axis. That means its y component is 0 and its x component equals 28.5 N.
 
Okay, I finally got it! Thanks for the help. :)