Recent content by BurpHa

Thank you! I think this is what I need.

But the friction force depends on the velocity, which I do not know how to formulate.

I could understand the problem perfectly; however, I do not know how to construct the problem. The problem states that two forces are acting simultaneously on the object, but how could I represent that fact mathematically? I really want to solve it, but I am facing this roadblock, so please...

I finally got this point. After reviewing the thread and my study materials, I could nearly understand the matter. Thank you anyone who has lent me a hand in rewiring my thinking 🤞✌💪💪

The equation is ##9.8 * 0.3 * m_A##.

I mean that when the static friction force on block A equal to ##\frac {19.6} {x + 2} * x##, the tension force on block A is still larger and therefore be able to win over the maximum static friction force.

Wait, I thought that the force on the right side (tension) is larger than the force on the left side (static friction.) That is why block A still moves forward with an acceleration.

You mean my image is not clear? Because other things are not related. ;)

Did you want to say the right side is always larger then the left side?

Alright, I have understood.

Ok, logically, it must be that the static friction force of block A equal to the force of gravity on block B, so mass of block A is: m_A * 9.8 * 0.30 = m_B * 9.8 m_A * 2.94 = 2 * 9.8 m_A * 2.94 = 19.6 m_A \approx 6.7 kg. However, when I look at block A individually, there is one thing...

Thanks you all for your help. As it turns out, I did not round off the numbers correctly, and I overlooked the meaning of significant figures.

I don't understand why the box could move when the applied force is equal to the static friction.

Less than.

I don't understand part (b) In part (a), I need to calculate the coefficient of the static friction: mg * \mu_static = 35 58.8 * \mu_static = 35 \mu_static = 35 / 58.8 \approx 0.6 So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot...