A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Please help! (:
Two objects have different weights. The first is three times as heavy as the second. If the objects are stacked (with the second block on the bottom and placed on a surface that exhibits friction), how much would the friction be? (Select the BEST answer).
a) Same as block 1 alone
b) Same as block 2 alone
c) Three times the amount of the second block alone
d) 8 times the amount of second block alone
e) None of the above
 2 years ago
Please help! (: Two objects have different weights. The first is three times as heavy as the second. If the objects are stacked (with the second block on the bottom and placed on a surface that exhibits friction), how much would the friction be? (Select the BEST answer). a) Same as block 1 alone b) Same as block 2 alone c) Three times the amount of the second block alone d) 8 times the amount of second block alone e) None of the above

This Question is Closed

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0If the friction depends on the total mass "pushing down", then the friction from the combination of blocks is proportional to the total mass "4x"

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0So E, none of the above.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0E is what it appears to me. You have to be a little careful to make sure they don't sneak an answer in that might work, but it just comes down to comparing 4x versus each of the other single blocks.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0Now if 4x = 1,451 pounds, what would the friction force being exerted on the box be? Or how would I find it?

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Is the total mass of the two blocks equal to 1451 pounds?

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Like your other problem, Force of friction = "mu" times the normal force.\[F _{friction} = \mu N\]

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0So the normal force is the force pressing up against the bottom of the blocks... since they gave it to you in pounds, that is a measure of mass times acceleration already... in other words, the normal force is equal to the weight, 1451 pounds.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Were you given a coefficient of friction, μ?

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0They said it is being pushed across a floor at a constant velocity.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0It is already moving as a constant velocity.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0ok, but that doesn't tell you much about the friction force.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0I'm just going to type the entire question out.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0Unless they tell you what the force that is pushing it horizontally is.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0A mover is pushing a 1,451 pound box across a garage floor. Assuming that it is already moving at a constant velocity, what is the friction being exerted on the box?

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0That's not enough to be able to solve and get a number, but you can solve in terms of that coefficient of friction. To see why, realize that the garage floor could be almost as slick as ice... so very little friction exerted, and the mover just pushes lightly to maintain constant velocity.... OR.... a very rough floor, requiring the mover to push hard to maintain constant velocity against the much larger force of friction exerted. The problem gives you no way to know the amount of roughness... that's what the μ is.

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0Dang... my physics prof never explains anything, so I'm not surprised that his exams are flawed.

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0So, the mover just has to push equal to the force of friction exerted, because as the problem says, the box is at constant velocity (i.e. no acceleration, therefore no net horizontal force).

pathosdebater
 2 years ago
Best ResponseYou've already chosen the best response.0Can you figure out what force must the mover push in order to keep the box moving at a constant velocity? (Probably not) Or, How much force does the mover have to exert to start moving the box?

JakeV8
 2 years ago
Best ResponseYou've already chosen the best response.0To keep it moving at constant velocity, the mover's push must equal the force of friction which is mu * N, or here, 1451 * mu, but since you don't know mu, you can't solve fully. I can't remember how to handle the static versus dynamic friction part... you have to push harder to start, due to static coefficient of friction, then it becomes easier after it's moving, but I don't recall how to set up the problem (sorry!!)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.