The tension in the cord in this system

In summary, the blocks A and B are sliding on the floor with a horizontal force F. The coefficient of kinetic friction between the two blocks is μ1 = 0.500 and between block A and the floor is μ2 = 0.250. The tension in the cord is F=m.a.
  • #1
Blueberrymuffim
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Homework Statement


A force F is applied horizontally to a block A of mass mA =
0.600 kg so that it slides on the floor with acceleration g/3.
Block B of mass mB = 0.200 kg rests on top of A, but is kept
in place by a cord which is tied to the wall at 45.0°. The
coefficient of kinetic friction between the two blocks is μ1 =
0.500 and between block A and the floor is μ2 = 0.250.
(a) Draw a free-body diagram for each block.
(b) Find the tension in the cord.
(c) Find the normal force that block A exerts on block B.
(d) Find the normal force that the floor exerts on block A.
(e) Find the force F.

Homework Equations


F=m.a fk = Mk*n

The Attempt at a Solution


I tried to find the Tension by the components of the vector. Initially, I found that since the angle of the force is 45 degree, then I have an isosceles triangle for the components which each force will be the same in the component by in different direction which changes the sign (just assuming).

Then, I separate the forces of the Tension in components Fy and Fx acting in the block B.

Fy= S (force of the component y of tension) + n (normal force) - W (mass of B * gravity)
I assumed that Fy would be equal to zero because it is not moving in the vertical direction which means the sum of all the components of this net force will be equal to zero.
So, I decided to isolate S + n= 1.96.
So, S+n = weight.
n=1.96 N - S

Fx= fb (friction force) + S(force of the component x of tension).
Fx= Mk*n+S
Fx=0.5n - S
Fx = 0.5(1.96 - S) -S
Fx=0,98 - 0.5S - S
Fx = m.a by the given information we have Fx= 0.653N
S = -3.266 N
 

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  • #2
I am confused about the letter b. I guess I can find the normal force of each blocks after I found the Tension of the cord.
 
  • #3
Have you tried (a) yet?
When you do that, you can derive equations from that to answer the other parts in order.
 
  • #4
Blueberrymuffim said:
S + n= 1.96.
A very good habit to get into is keeping everything algebraic, only plugging in numbers at the very end. It has many advantages.
Blueberrymuffim said:
Fx = m.a by the given information we have Fx= 0.653N
I thought we were dealing with block B here.
 
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  • #5
Merlin3189 said:
Have you tried (a) yet?
When you do that, you can derive equations from that to answer the other parts in order.
Yes, I did the free body diagram. It is attached here.
Honestly, I haven't used any equations on it yet. My initial thought was to think about Force net and tension and they are the sum of the forces acting in the system, right? Obviously, it also uses the second law of Newton F=ma, but I don't really see how to get on this step yet.
 

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  • #6
haruspex said:
A very good habit to get into is keeping everything algebraic, only plugging in numbers at the very end. It has many advantages.

Can you be more specific on this, please? I've been trying to look both blocks and find things in common in a way that I may end with substitutions and I can find common variables.

I thought we were dealing with block B here.
Yes, we're dealing with block B here. Weight of B is 1.96N. and what you're quoting is my initial idea of components of force net which would be S+n-W. Since the only variable I have is the weight s+n = 196N
 
  • #7
Blueberrymuffim said:
Yes, we're dealing with block B here.
... which is stationary... so why the nonzero ma?
 
  • #8
From your diagram for block B, you should be able to find (b) and (c) without reference to block A at all.

For block A, you don't seem to have shown any vertical force from B. Think about Newtons 3rd law for the force n1 .
I presume n2 is the force from the floor pushing A upwards. The 3rd law force acts on the floor (Earth).
You show the force from the gravity of Earth pulling A downwards. The 3rd law force acts on the Earth.

Edit: PS. Do you remember sin(45)=cos(45)

Edit2: PPS. You seem to use force symbols which do not appear on your diagrams (S, W, Fy ). IMO this is unhelpful. I think you can mark on all the forces you need and use only those in your equations.
 
Last edited:

Related to The tension in the cord in this system

1. What is tension in a cord?

Tension in a cord refers to the pulling or stretching force applied to the cord. It is the force that resists any changes in length or shape of the cord.

2. How is tension in a cord measured?

Tension in a cord is typically measured in units of force, such as Newtons (N) or pounds (lb). It can be measured using a tension meter or by calculating the force needed to stretch the cord a certain distance.

3. What factors affect the tension in a cord?

The tension in a cord is affected by several factors, including the weight of the objects attached to the cord, the angle at which the cord is pulled, and the properties of the cord itself (such as its length, thickness, and elasticity).

4. Why is tension important in this system?

Tension is important in this system because it determines the stability and strength of the cord. If the tension is too low, the cord may break or slip, while if it is too high, the cord may become stretched or damaged.

5. How can tension in a cord be adjusted?

Tension in a cord can be adjusted by changing the force pulling on the cord, adjusting the angle at which the cord is pulled, or using a different type of cord with different properties. It is important to carefully consider the desired tension in a system to ensure safety and effectiveness.

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