Solving Pulley & Block Problems: Acceleration of A

[chenying]

Homework Statement

Body A in Fig. 6-36 weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are μs = .56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x-axis be down the slope. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

(see the attachment)

Homework Equations

$$\Sigma$$F = ma

Frictional force = F$$_{k}$$ = $$\mu$$$$_{k}$$ * F$$_{N}$$

T = tension in string

The Attempt at a Solution

for a, I know the answer is 0 because the block is at rest, therefore it will remain at rest until another force acts on it.

Now for b, I stated that the sum of all forces on the block would equal m*-a, a being negative because the problem stated the coordinate axes. So

$$\Sigma$$F = ma

mgsin$$\Theta$$ - F$$_{s}$$ + T = m*-a

the F$$_{s}$$ should be F$$_{k}$$

plugging the numbers in and solving for a, i get a value of 7.5 /s^2, which is wayyyyy too fast. I do not know what is wrong with my set up, so can anyone help me out?

 

[ehild]

When the body A moves upwards (negative direction) the friction points downwards as it acts against the motion. The tension is negative as its acts uphill. The resultant force points downwards, and so does the acceleration: it is positive.
Do not assume sign for the acceleration, you will get it from the solution. Newton's second law states F=ma, both F and a vectors. ehild

 

[tomcue]

Hello,

Thank you for sharing your attempt at solving this problem. It seems like you have a good understanding of the relevant equations and concepts.

First, let's address your solution for part (a). You are correct in stating that the acceleration of body A will be 0 since there are no forces acting on it. However, I would recommend writing out the equation explicitly to show this:

\SigmaF = ma

0 = 0 * a

Since there are no forces acting on the block, the net force is 0 and therefore the acceleration is also 0.

Now, let's move on to part (b). Your setup is mostly correct, but there are a few things that need to be adjusted. First, the frictional force should indeed be F_{k}, not F_{s}. This is because the block is moving, so the kinetic friction coefficient should be used.

Next, when you write out the equation for the sum of forces, make sure to include the direction of each force. In this case, the force of gravity is acting down the incline, so it should be written as -mgsin\Theta. The tension in the string, T, is also acting down the incline, so it should also be written as -T. The kinetic friction force, F_{k}, is acting up the incline, so it should be written as +\mu_{k} * F_{N}. The normal force, F_{N}, is perpendicular to the incline, so it should be written as mgcos\Theta.

Putting all of this together, the equation should look like this:

\SigmaF = ma

-mgsin\Theta + \mu_{k} * mgcos\Theta + (-T) = m * a

Now, you can solve for the acceleration, which should be a = 2.2 m/s^2.

For part (c), the setup is similar to part (b) except that the block is now moving in the opposite direction, so the signs of the forces will be different. The equation should look like this:

\SigmaF = ma

mgsin\Theta + \mu_{k} * mgcos\Theta + (-T) = m * a

Solving for the acceleration, you should get a = -3.3 m/s^2.

I hope this helps to clarify the setup and solution for this problem. Remember to always carefully consider the direction