Question

You may be familiar with Newton's Second Law of Motion, SF = mā. In English, this equation says: The sum, or net, (S) of the forces (F) acting upon an object equals (=) the mass (m) of the object multiplied by the object's acceleration (ā). Even if you are familiar with this famous equation, did you notice the arrows above F and a before? If so, you may have wondered what they mean. They are vectors! This means that they have both a magnitude and a direction and can be expressed in terms of their components. These arrows are a flashing red light that alerts you to the fact that these quantities imply two equations; one in the x direction and one in the y direction. A pack of four Arctic wolves are exerting four different forces upon the carcass of a 500 kg dead polar bear. Wolf one is pulling on the bear with a force of 20 Newtons (N) due north. Wolf two is pulling on the bear with a force of 30 N, 30 degrees north of west. Wolf three is pulling on the bear with a force of 25 N due west, and wolf four is pulling on the bear with a force of 35 N, 10 degrees west of south. Determine the net force acting upon the polar bear and the direction of that force measured north of west. Then compute the magnitude of the acceleration of the polar bear. Wow. This much information can be scary at first but we are just going to use the techniques we have learned above and work our way through. Take a deep breath. You will be fine. Your first step should always be to draw a diagram of the situation. Do so here: (Hint: Draw this as though you are looking down on it from directly above with the bear at the origin of your coordinate system, like a two-dimensional tug-o'-war.

I would really appreciate it if someone can help me with the whole question. thank you.

Answer 1

LON F2 =30N MW W t=25N. Ho Fu = 35nt S Wolf one Exi 20 N Wolf Two - 25.981N FE Fy Zox Gs 30 15 N Wolf Three Ex -250 Wolf Four Fx = 35x sin loo -6.077N 34.468 m

Add all - - components, Ex = 0 25.981-25-6.077 Exa 57.058.. Add up all y components so t 15 to-34.488 Fy 0.533 N FE Fot. Fy 57.1 Total force lo 0.5339° a m 57.7 500 1149 m/s²