Question

Question 4 (force vectors) Two people are pulling on a crate through two ropes. Person A is on the left pulling toward the left with a force of 50 N at 45° and person B is on the right pulling toward the right at an angle of 30°. Here is what the situation looks like from above. The resulting acceleration of the crate is at 0°. acceleration direction person B F 30° person A 50 N 45° 130% 45% crate a. What is the magnitude of the force from person B on the crate? b. What is the angle of the force from person A on the crate? If a person C comes in and starts to push the crate to the left from the right with 10 N of force, acceleration direction person B F @ 30° person A . 50 N 45; 130 45° person C 10 N crate c. Can person B adjust the value of F to keep the crate accelerating in the same direction? If so, how much force must F be?

Answer 1

(b) the angle of the force from person on the crate with Horizontal. Hence, the angle of the force from person A on the crate with Horizontal. Is = (135 degrees) (c) for the crate acceleration the same direction than Sigma f_x = infinity So sin 45 + 10 = (R) sin 30 [(50 times 1/squareroot 2) + 10] times (2) = (R) (F) = 90.71051 Ans \r\n (A) for the Resultant acceleration as 0 degrees C than sigma f_x = 0 & sigma f_y = 0 50 so sin 45 = (F) sin 30 so times 1/squareroot 2 times (2) = (F) (A) (F) = 70.71 H