Homework Statement
http://imgur.com/BPPbwpu
Homework Equations
s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation.
The Attempt at a Solution
Let u be the velocity at which the projectile was launched.
It is given that sx = symax. So...
O.K., I realized I grossly overcomplicated things.
Since the force of static friction on Block 1 due to Block 2 is the only force in the x-direction acting on Block 1, and Block 1 has acceleration a, then Newton's Second Law states that:
∑ Fx = m1*a = fs12
Thus, the force of static friction...
Homework Statement
http://imgur.com/C0XYEKw
Homework Equations
Fnet = m*a
FG = m*g
fstatic = μ*N
The Attempt at a Solution
If both Block 1 and Block 2 have acceleration a in the x-direction and have an acceleration of 0 in the y-direction, then the force of static friction on Block 1 due to...
Since the workers are pulling with a force f on the rope, then due to Newton's Third Law, the rope would exert a force f on the workers in the opposite direction, so I am missing another 2 f forces, right?
So the acceleration should be:
a = (4*f - (m + m + 0.5*m)*g)/(m + m + 0.5*m)
Hey, are you from Australia? I remember doing E.E.I.s... God I hated them. Well done doing quantum physics as a topic for your E.E.I. though. Complex and challenging, right? That sounds like a hard topic to do not only because the theory is difficult but it is difficult to do an experiment on (I...
Homework Statement
http://imgur.com/SE50jeP
Homework Equations
Fnet = m*a
FG = m*g
The Attempt at a Solution
I'm so confused about this problem and I still can't solve it after weeks.
The initial distance of the workers is irrelevant. If I treat the workers and the platform as a single...
Rather, something like this:
If a ≠ 0 and b ≠ 0, then if a = b, then 1/a = 1/b.
Well, I'm probably not using the inequality sign correctly and trying to use equality rules for the "greater than" sign. But basically, if x > 0, then a "=" x, and b "=" 0, and using the algebra "rule" above, if x...
Oh, well, there is nothing wrong, then. I thought that rule was wrong because if x > 0, then dividing 1 by both sides of the inequality yields 1/x > 1/0. My concern was that 1/0 is not defined. There is clearly something wrong with the algebra, but what is wrong?
Thank you.
If x approaches 0 from the right, then 1/x will become arbitrarily (positively) large, which is definitely still greater than 0. Hm... I still don't see what could be wrong.
Before I posted, I Googled "inequality properties" and I found a popular school study guide called "Sparknotes", which...
If a > b, where a > 0 and b > 0, then 1/a < 1/b.
But what if b = 0? For example, if x > 0, meaning if x is a positive number, then it should be that 1/x > 0.
Yes, yes, I know I would be dividing by 0, but it doesn't make sense intuitively. If x is a positive number, then obviously 1/x is a...
I know that the reduced mass, μ, of an object is:
\mu = \frac{1}{\frac{1}{m_1} + \frac{1}{m_2}}
\mu = \frac{m_1 m_2}{ m_1 + m_2 }
But is there a general formula (or a simplified expression) for finding the value of:
\frac{1}{ \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n} } ?
Thank you.