anonymous
  • anonymous
Two people pull on a stubborn mule. One person pulls at a 70 degree angle with 50 lbs of force, and the other person pulls at a 100 degree angle with 40 lbs of force. Find the single force that is equivalent to the two forces. Find a third force that a person would need to exert to make the resultant force equal to zero. Explain.
Physics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
First, we need to take the dot product of the two vectors. \[F_1 \cdot F_2 = |F_1||F_2| \cos(\theta_2 - \theta_1)\] Second, we need to find the components of each force and add them. \[F_x = F_1 \cos(\theta_1) + F_2 \cos(\theta_2)\]\[F_y = F_1 \sin(\theta_1) + F_2 \sin(\theta_2)\]The third force must be equal but opposite to these. The angle at which this force acts\[\tan^{-1} \left( \left \vert F_y \over F_x \right \vert \right )\]
mathmate
  • mathmate
|dw:1327985297401:dw| Case A: F1=50, a1=70 degrees F2=40, a2=100 degrees. All angles measured with respect to the x-axis. Equivalent force has the same x- and y-components, so Fx = 50cos(70)+40cos(100) Fy = 50sin(70)+40sin(100) angle of equivalent force = arctan(Fy/Fx) magnitude of equivalent force = sqrt(Fx^2+Fy^2) Case B is similar to case A, with the angle 100 degrees replaced by -100 degrees.

Looking for something else?

Not the answer you are looking for? Search for more explanations.