Block A, with a mass of 10 kg, rests on a 30° incline. The coefficient of kinetic friction is 0.20. The attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. Block B, with a mass of 3.0 kg, is attached to the dangling end of the string. The acceleration of B is:
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
I've already got the following forces for block A which are:
Fg parallel = 49 N
Fg perpendicular = 49 sq. root of 3N
Fg =98 N
Ff = 16.97 N
I can't seem to get the Tension Force and the acceleration of block A so that I can compute for the acceleration of Block A since the tension force of Block A and B are the same.
Not the answer you are looking for? Search for more explanations.
What I usually do with these problems is I would isolate A and compute for the force acting on it, and then do the same with B.
I like to visualise the question. When a block is on a slope, there would a be a force pulling down the slope. (Remember, isolate the system! Don't think about B!)
The other forces that would act on it is the tension from the string and the force of friction.
We know that with respect to weight, the force that would be "pulling down the slope" is \[-m_Ag \sin \theta\]
It's a negative because it's going to the left.
Then we have friction force, given by\[f_k=\mu_kN\]
Wherein N is the normal force
We found from our drawing that \[N=m_agcos \theta\]
The last force would then be T, which I would leave on its own since we've yet to find it.
So the net force on A would be\[F_A = m_a(a) = -m_agsin \theta-f_k+T\]
Now, let's look at B: (Isolate the system!)
1. The forces that would act on be would be the tension pulling up represent by T.
2. The weight of the block B. F=m_b(g)
So the net force on B would be
\[-T+m_bg = m_ba\]
So now you have two equations with two unknowns. You don't know T, and you don't know a. But you know that the tension pulling at A would be the same as that pulling at B, just different directions. So you can equate the two equations together, plug in your values, and voila!
Why is the Tension force of block b negative and the F of the weight positive? :)
Oh I see, thank you so much ^_^
is a = -2.81 m/s^2 ?
Nope! Slowly calculate.
There's a lot of numbers involved haha!
With regards to your why is the weight positive and the T negative question:
Forget what I said earlier hahaha
The weight will be GREATER than T because the block is going DOWN.
That means that
\[W - T = F_b\]
is a = 0.57 m/s^2? :)))
It should be a negative number, I believe.
My professor said that it should be 0.20 m/s^2, up :(
Perhaps, I've yet to calculate it. Do persevere :)