Finding Tension, Acceleration and Net Force?

Question

Two blocks (A and B) are tied together and are being pulled along horizontally by a rope attached to Block A with a force of 84N. Block A is 35kg with a coeifficient of friction of .02. Block B is 16kg with a coefficient of friction of .03. Find the following.

What is the accelration of either block?
What is the net Force on Block A?
What is the Tension on the rope between the two blocks?

16kg T=? 35kg
[B]----------------[A]------- > 84N
u=.03 u=.02

Answer 1

The force of friction on block A is
Ff = Fn * u
= 35 kg * 9.8 N/kg * 0.02

The force of friction on block B is:
= 16 kg * 9.8 N/kg * 0.03

subtract these forces from the 84 N to get the net force on A&B
This total net force goes toward accelerating the blocks.

Use F = ma to find the acceleration of A&B (the same for both).
a = (84N -Ffa - Ffb) / (m_a + m_b)

Then knowing 'a', use F=ma again to find the two different forces on A, B due to acceleration.
Fa = m_a * a
Fb = m_b * a

To find the net Force on A, and the tension, use your free body diagram and draw in all the forces. Find the net force by adding up all the forces acting on the object that go to the right, and subtract the forces going to the left.

Answer 2

For style a million you began out accurate by technique of using breaking the pressure into x and y. You made the mistakes pondering the Fn is altered by technique of using the pressure (it isn't). Fn = Fgy (rigidity of gravity in the Y route). The rigidity of hysteria is a mixture of each and every x & y guidelines going up the incline. extra frequently than now to not sparkling up it you separate the finished forces into the x and y instructions (Fn and Friction). then you actually definately seem at Ftx - Fgx - Ffx = ma (Ffx rigidity of friction in x course). Do the equivalent for the y route to seek out Fty and also then you definately ft^2 = Ftx^2 + Fty^2. style 2b you had perfect. provided that the helium replaced into miraculous pressure up and the load replaced into miraculous pressure down. pondering the actual incontrovertible truth that there replaced into once no action the pressure up equals the pressure down, subsequently the information superhighway pressure on the rope is the same as 0. i believe it truly is optimal, expectantly it helped.

Answer 3

First we calculate the force of friction acting on A & B:

f=uN where f= the force of friction, u the coefficient
of friction, and N the weight of the block

Substitute:

f on A:

f=0.02*35*9.8=6.86N

f on B:

f=0.03*16*9.8=4.70N

Total f on A & B =6.86+4.70=11.56N

Net force=84-11.56=72.44N

Compute for a using the formula F=Ma, where F
is the net force acting on both blocks, M the mass
of both blocks, and a the acceleration.

Substitute the given values:

72.44=(34+16)a
a=72.44/50=1.45m/s^2 (acceleration on either block)

Compute the tension on the rope,T.

--------->F=T-4.70
a=1.45
[B]----->T
<----4.70

F=Ma
F=16*1.45=23.2N

F=T-4.70
T=23.2+4.70=27.9N


Compute the net force on A:

------->Net force, F
27.9 <----[A]------>84
<-----f=6.86

F=84-27.9-6.86=49.24N

Check your answer: the net force acting on A + the net force acting on B= the total net force acting on A&B: 49.24+23.2=72.44N