What Is the Acceleration of Block A and the Tension in the Cord?

[nothingkwt]

Homework Statement

Block A has a weight of 400N and block B has a weight of 100N. The coefficient of friction between all surfaces of contact are μ_s = 0.7 and μ_k = 0.2.
Knowing that θ = 60°, determine the acceleration of block A and the tension in the cord. Assume block A is moving downwards.
I know I should be solving for the acceleration but I figured if I get the tension I could use ∑F = ma on block B to get the acceleration there and that would be the same acceleration on block A.

Homework Equations

∑F = ma

The Attempt at a Solution

(BLOCK A)
took the inclined plane and it's normal as the axis and used Newton's equation on the perpindicular axis to get the normal (N - Wcosθ = 0) and got N = 200N. Then used the plane's axis at the moment the tension force equals the static friction (Wsinθ - 2T - μ_sN = 0) but I got 103.7N as the answer when it should be 127N and I don't know what my mistake is.

 

[Simon Bridge]

determine the acceleration of block A and the tension in the cord. Assume block A is moving downwards. ... used the plane's axis at the moment the tension force equals the static friction

... why static friction?
... how did you account for the mechanical advantage of the pulley?
... how did you incorporate the action of block B.

 

[nothingkwt]

... why static friction?
... how did you account for the mechanical advantage of the pulley?
... how did you incorporate the action of block B.

I was trying to get the tension on the cord for before the block started moving

I didn't incorporate block B yet I thought I could get the tension in the cord from block A .

 

[dauto]

Block B and Block A have different accelerations.

 

[Simon Bridge]

I was trying to get the tension on the cord for before the block started moving...

How does this help you?
Are you not asked for the situation that the block is moving?

I didn't incorporate block B yet I thought I could get the tension in the cord from block A .

Don't you think that the weight of block B would have some impact on the size of the tension?

In other words:
You should revisit those decisions.
You have correctly surmised that you canot get the tension without the acceleration just by considering block A. But choosing to work out the tension when the blocks are balanced is not going to help much. The tension when stationary is not the same as the tension when moving. This means you have to use block B to help find the tension.

This is just like other problems you have done - you will end up having to solve simultaneous equations. You will need two free body diagrams - write out the equations for each.
(I am guessing that the two pulleys can be considered frictionless?)

Note:
If A moves a distance x down the slope, how far does B rise?
If A has instantaneous velocity v down the slope, what is the velocity of B?
How are their accelerations related?

 

[Satvik Pandey]

Homework Statement

Block A has a weight of 400N and block B has a weight of 100N. The coefficient of friction between all surfaces of contact are μ_s = 0.7 and μ_k = 0.2.
Knowing that θ = 60°, determine the acceleration of block A and the tension in the cord. Assume block A is moving downwards.
I know I should be solving for the acceleration but I figured if I get the tension I could use ∑F = ma on block B to get the acceleration there and that would be the same acceleration on block A.

Homework Equations

∑F = ma

The Attempt at a Solution

(BLOCK A)
took the inclined plane and it's normal as the axis and used Newton's equation on the perpindicular axis to get the normal (N - Wcosθ = 0) and got N = 200N. Then used the plane's axis at the moment the tension force equals the static friction (Wsinθ - 2T - μ_sN = 0) but I got 103.7N as the answer when it should be 127N and I don't know what my mistake is.

(Wsinθ - 2T - μ_sN is not equal to 0.It should be equal to Mass of block * acceleration of Wa, otherwise no net force will act on it and the block will be in equilibrium.When you draw the FBD of two block you will get 2 equations.Also the acceleration of Wb is twice the acceleration of Wa (third equation). Solve the three equations and get the answer.
But there is still an easier way, you remove the inclined plane and assume (Wa cos\theta-\muR) is acting on it instead of mg (as in normal conditions).

http://www.oldschool.com.sg/modpub/20204469354a2bdb19c0115
Convert your figure into this figure and assume force acting on load is (Wa cos\theta-\muR) in downward direction (this is not the net force on Wa) and do the further calculations.

 

[Simon Bridge]

Also:... the angle being a nice number means you can just substitute the fractions for the sine and cosine.
sin(60)=sqrt(3)/2, cos(60)=1/2

 

[nothingkwt]

Also:... the angle being a nice number means you can just substitute the fractions for the sine and cosine.
sin(60)=sqrt(3)/2, cos(60)=1/2

∑F = ma


m_a*g*sin(60) - 2T - μ_k * N = M_aa_a

(for block A and about the plane's axis)

Where T = m_bg + m_ba_b (from block B)


then I took the datum at the pulley that is on the ramp and got length = 2 S_a + S_b

deriving with respect to time twice I get . . . 0 = 2a_a + a_b

Substituting in the equation I get 1 unknown but still got an answer of a_a = 1.739 m/s² and that is not the right answer (a_a = 1.36 m/s²)

Is it because the length of the line equation is wrong since the cord is at an angle?

 

[dauto]

The formula 0 = 2a_a + a_b uses a sign convention for the accelerations different than the convention adopted by the other equations.

 

[nothingkwt]

The formula 0 = 2a_a + a_b uses a sign convention for the accelerations different than the convention adopted by the other equations.

Thank you that was it. I now got the right answer.