At a state fair truck pull, two pickup trucks are attached to the back end of a monster truck as illustrated in the figure. One of the pickups pulls with a force of 1600 pounds and the other pulls with a force of 3200 pounds with an angle of 45° between them. With how much force must the monster truck pull in order to remain unmoved? [HINT: Find the resultant force of the two trucks].

Thank you for helping me set this up Reiny.

Fr = 1600[0o] + 3200[45o] =

1600 + 2263+2263i = 3863 + 2263i =
4477N[30.4o] N. of E.

To find the force with which the monster truck must pull to remain unmoved, we first need to find the resultant force of the two pickup trucks.

The force exerted by the first pickup truck is 1600 pounds. Since this force is only in one direction (directly opposite to the motion), we can consider it as a negative force.

The force exerted by the second pickup truck is 3200 pounds, and it is at an angle of 45° with the horizontal. We can resolve this force into two components: one in the direction of motion (parallel to the negative force) and one perpendicular to it.

The parallel component of the force is given by: parallel force = force * cos(angle)
parallel force = 3200 pounds * cos(45°)
parallel force ≈ 3200 pounds * 0.707 (cosine of 45° is approximately 0.707)
parallel force ≈ 2264 pounds

The perpendicular component of the force is given by: perpendicular force = force * sin(angle)
perpendicular force = 3200 pounds * sin(45°)
perpendicular force ≈ 3200 pounds * 0.707 (sine of 45° is approximately 0.707)
perpendicular force ≈ 2264 pounds

Now, the two pickup trucks have forces of 1600 pounds (negative direction) and 2264 pounds (positive direction). To find the resultant force, we subtract the negative force from the positive force.

Resultant force = 2264 pounds - 1600 pounds
Resultant force = 664 pounds

Therefore, the monster truck must pull with a force of 664 pounds in order to remain unmoved.

Can't see your diagram, but it looks like a straightforward cosine law

R^2 = 1600^2 + 3200^2 - 2(1600)(3200)cos 135°
R = ....