anonymous
  • anonymous
You push against a steamer trunk with a force of 800 N at an angle alpha with the horizontal . The trunk is on a flat floor and the coefficient of static friction between the trunk and floor is 0.55. The mass of the trunk is 87 kg. What is the largest value of alpha that will allow you to move the trunk?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
the situation of your problem is described by the subsequent drawing: |dw:1438333970203:dw|
Michele_Laino
  • Michele_Laino
the pressure on the floor has the subsequent magnitude: \[\Large mg + F\sin \theta \] whereas the driving force, has the subsequent magnitude: \[\Large F\cos \theta \] the trunk will move, if and only if the subsequent condition is checked: \[\Large F\cos \theta > \mu \left( {mg + F\sin \theta } \right)\] where \mu is the coefficient of static friction
Michele_Laino
  • Michele_Laino
hint: after a simplification, we can write: \[\Large \cos \theta - \mu \sin \theta > \frac{{\mu mg}}{F}\] if we divide last condition by: \[\Large \sqrt {{\mu ^2} + 1} \] we get: \[\Large \frac{1}{{\sqrt {{\mu ^2} + 1} }}\cos \theta - \frac{\mu }{{\sqrt {{\mu ^2} + 1} }}\sin \theta > \frac{1}{{\sqrt {{\mu ^2} + 1} }}\frac{{\mu mg}}{F}\] which can be rewritten as follows: \[\Large \cos \left( {\theta - \varphi } \right) > \frac{1}{{\sqrt {{\mu ^2} + 1} }}\frac{{\mu mg}}{F}\] where \phi is such that: \[\Large \tan \varphi = \mu \]

Looking for something else?

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

More answers

Michele_Laino
  • Michele_Laino
oops.. I have made a typo, here is the right formula: \[\Large \cos \left( {\theta + \varphi } \right) > \frac{1}{{\sqrt {{\mu ^2} + 1} }}\frac{{\mu mg}}{F}\]

Looking for something else?

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