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.
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?
2 answers
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?
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?
0 answers