One more tricky 2nd Newtonian Law with friction problem

[frankfjf]

This time I'm uncertain as to how to start:

Body A in Fig. 6-34 weighs 98 N, and body B weighs 70 N. The coefficients of friction between A and the incline are ms = 0.50 and mk = 0.27. Angle is 48°. 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?

I'm sorry I can't produce the figure but it seems to be your standard two blocks tugging on each other on an inclined plane situation.

I figured the answer to a would be 0, since an object at rest stays at rest according to Newton, and got that one right intuitively.

However, I'm uncertain how to proceed for b and c.

I attempted to setup equations for both bodies.

For body A, I've got:

X component: f - wA(weight of body A) - Tcos(theta) = ma.
Y component: Fn + Tsin(theta) - wA = ma

For body B, I've got:

T - wB = ma

Can I proceed like this or is something off?

 

[topsquark]

This time I'm uncertain as to how to start:

Body A in Fig. 6-34 weighs 98 N, and body B weighs 70 N. The coefficients of friction between A and the incline are ms = 0.50 and mk = 0.27. Angle is 48°. 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?

I'm sorry I can't produce the figure but it seems to be your standard two blocks tugging on each other on an inclined plane situation.

I figured the answer to a would be 0, since an object at rest stays at rest according to Newton, and got that one right intuitively.

However, I'm uncertain how to proceed for b and c.

I attempted to setup equations for both bodies.

For body A, I've got:

X component: f - wA(weight of body A) - Tcos(theta) = ma.
Y component: Fn + Tsin(theta) - wA = ma

For body B, I've got:

T - wB = ma

Can I proceed like this or is something off?

I presume that B is hanging?

Your answer to a may not be correct.
Just because we started A at rest doesn't mean that it will stay at rest because there is possibly a net force acting on it, the unbalanced part coming from mass B.

I'll sketch b and c first, then get back to a.
Draw a FBD for each mass. For mass B do yourself a favor and choose a positive direction upward...it makes sense with the positive down the slope you've chosen for mass A. Now, what direction for the friction force on A? Kinetic friction always acts in the opposite direction of the motion, so the direction of motion of A tells you what direction kinetic friction will be. So set up your equations. Warning: Make sure when you do the net force on each mass that the mass you are using is the mass of the object in the FBD...for example on mass B, you've got a tension (T) upward and a weight (mBg) downward, so Newton's Law says: $$F_{net}=T-m_Bg = m_Ba$$. On block A, $$F_{net}=m_Aa$$.

Now, for part a we have to be a bit more careful. We are starting A stationary and the question is will it move? I think the best way to answer this is to assume it will not and find out what static friction force is required to keep it stationary. So set up your FBD for mass A. What direction do we assign for the static friction? The answer is: GUESS! We don't know what direction to assign off the top of our heads, so just pick one. If you choose the wrong direction, the sign of the friction force you work out will simply be negative.

So how do we tell if its moving? Calculate the maximum static friction force using ms and compare your maximum with your calculated value. Obviously if fsmax is greater than the friction force you calculated it won't move. If not then mass A will, in fact, be moving. So then you have to go back to the start of problem a and use kinetic friction this time.

It's a long process, but once you've done part b you've seen the whole process so you can (carefully!) take some shortcuts with the equations for a and c.

-Dan

 

[kyle8921]

I would first start by analyzing the problem and understanding the given information. It is important to note that this is a 2nd Newtonian Law problem, which means that we will be using the equation F=ma to solve for the acceleration of Body A.

To start, we can break down the forces acting on Body A and Body B. For Body A, we have the force of friction (f), the weight (wA), and the tension (T). For Body B, we have the tension (T) and the weight (wB).

Next, we can use the given information to calculate the force of friction (f) using the equation f=μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force would be equal to the weight of the body, so we can calculate f for both cases (moving up and down the incline).

For part (b), when Body A is moving up the incline, the friction force (f) will be acting against the direction of motion. This means that the equation for the x-component of the forces on Body A would be: f - wA - Tcos(θ) = ma. The y-component equation would be: Fn + Tsin(θ) - wA = 0, since the body is not accelerating in the y-direction.

For part (c), when Body A is moving down the incline, the friction force (f) will be acting in the direction of motion. This means that the equation for the x-component of the forces on Body A would be: f + wA - Tcos(θ) = ma. The y-component equation would be: Fn + Tsin(θ) - wA = ma, since the body is accelerating in the y-direction due to the incline.

We can then solve for the acceleration (a) using the equation F=ma, where F is the net force acting on the body. It is important to note that the tension (T) will cancel out in both cases, so we can solve for a using only the x-component equation.

In summary, to solve this problem, we need to first analyze the forces acting on both bodies and then use the given information to calculate the force of friction. From there, we can use the x-component equation to solve for the acceleration in both cases. I recommend double-checking your equations and calculations