Posted by Nilan on .
Two forces, F1 and F2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 69.0 N and F2 = 24.5 N. What is the horizontal acceleration (magnitude and direction) of the block? F1 is pointing at the top left corner of the box forming a 65 degree angle with the parallel of the top of the box and the corner, F2 is pointing at the middle of the right side of the box.
2 questions: Do I subtract F1 (29.16) - F2 (24.5) Or do I subract F2 - F1? This would make a difference in a negative sign.
2) How do I determine then if the acceleration direction is right or left?
Physics Please Help - drwls, Saturday, January 21, 2012 at 12:51pm
Add the x components of F1 and F2. Then divide the result by the mass, 5 kg.
I would need to see the drawing to give you a more explicit answer. It does not matter what part of the block the vector arrow touches. It is the direction of the arrow that counts.
Physics Please Help - Nilan, Saturday, January 21, 2012 at 1:08pm
Force 1 is pushing on the top left corner of the box. The arrow forms a 65 degree angle with the verticle (as if the box were extended straight left, from 9:00 to the arrow = 65 degrees)
I found F1 horizontal component to be 69 cos 65 = 29.16 N
F2 is pushing to the left. F2 = 24.5 N
Am I right if I combine 29.16 - 24.5 (since it's pushing left) and divide by 5 to get 0.932 m/s^2 acceleration.
Then, since the F1 pushing right (29.16 horizontal component) is greater than the 24.5 pushing left, the direction is RIGHT. Is this correct?