III) Determine a formula for the magnitude of the force F → ex...

Question

III) Determine a formula for the magnitude of the force F → exerted on the large block () in Fig. 4–56 so that the mass m_A does not move relative to m_C. Ignore all friction. Assume m_B does not make contact with m_C .

Accepted Answer

Welcome back. Everyone. In this problem. A ball of mass M prime is attached by a massless string to the tip of an empty bottle of wine that is kept on a trolley as shown in the figure below the mass of the bottle is M that's a common M. Whereas the mass of the trolley itself is a capital ma force F is applied on the trolley towards the left in such a way that the trolley starts moving to the left while the bottle stays still with respect to the trolley. As shown assuming the mass M prime doesn't touch the trolley while the trolley moves, what would the expression of the force F be? And we are ignoring friction for or enter choices. A says that F is equal to the expression M prime plus M plus capital M all multiplied by GB says F equals capital M multiplied by M prime multiplied by G divided by the square root of the expression M squared minus M prime squared C says the force equals the expression M prime M plus capital M multiplied by M prime multiplied by G all divided by M squared minus M prime. Squared. And D says the force F is zero newtons. Now to make sense of our problem, OK. We're trying to figure out the, an expression for the force and for that expression before we figure out what the expression is, let's try to draw a free body diagram then to represent all the forces that are at play. OK. So here we have our trolley, OK? Of mass M and then we have our wine bottle on top of the trolley. OK. Let me look something like this. And then connected to our wine bottle, we have our mass M prime, OK? It is call that M prime here. So we have M, we have M prime and then we have our mass capital M OK? And then we have our force F that's uh pushing the charity. OK? We're acting on the trolley in this direction. OK? So that's F no, if we think about all our weights, OK. If we look at starting with our bottle with mass M, then its weight is acting downwards MG while a normal force is acting opposite to its weight upward. If we look at our mass and prime, OK, then M prime. The string that M is the string that is connected to M prime has a tension. OK? And it's called a tension T. And for that tension T, it has a vertical component and it also has a horizontal component. Now, for its vertical component, if we take theta as the anger that's formed between the string and the vertical, then its vertical component is going to be T cos theta while its horizontal component is going to be T sine theta. And of course, the mass M prime has its own weight acting downwards as M prime multiplied by G. And now, as we said, we're trying to find an expression for our force. And if we look back at our answer choices, it's in terms of, in terms of our three masses and G, OK. So first, let's ask ourselves, what do we know about the force and our motion here? Well, recall, OK, that by Newton's second law of motion, the sum of forces that are acting in a particular direction equals the mass multiplied by the acceleration. Now, here we have our force acting horizontally. So if we look at the sum of forces in the X direction, then what we're saying is that the force which in this case would be F is equal to the mass. Now, we're talking about the, the, the complete mass so that M prime, OK, plus M, the mass of the bottle plus capital M, the mass of the trolley multiplied by A. OK. So that would be our force. Now, here we have this expression, but it's in terms of the masses. And a remember we want to write it in terms of G. So we'll have to do some more work to try to figure out what forces are, are working here both for our, the mass of our uh what's this object that we call it again, the mass of our, well, our ball of mass M prime and the mass of our bottle M OK. So given that the components remain stationary relative to each other, and we are assuming the entire system accelerates at a rate of a, then let's try to first start by figuring all the forces on M. OK. So resolving, let me make a line here to show a separation, resolving the forces on M that is M prime. Sorry. First, in the vertical direction, we know that some of the forces are going to be equal to zero, there's no vertical acceleration. So with that in mind, that means the Y component taking upwards to be positive T cas theta is acting against M prime G. So that means T cas theta OK equals M prime G. And let's label that as equation one next in the X direction, we can use Newton's second law of motion that sum of forces equals the mass multiplied by the acceleration. Remember we assume the entire system accelerates at a rate of A. So that means the horizontal component of our attention T sine theta is going to be equal to M prime multiplied by A. OK. And let's call that one equation too. Now, what else do we, what else? What else can we figure out here? Well, let's consider the forces on our mass and the forces on our bottle. OK. So considering the forces on M, OK, then again, if we, if we look at M then for M the tension T, OK, that's in our string because our tension would also be acting here is going to be equal to the mass of the bottle. M multiplied by the acceleration. OK. So in other words, for him, T we write it here T equals ma. OK. So let's call that equation three. Now we have some formula with our acceleration a involved. So let's see if we can simplify to eventually solve for a notice from equations uh equations one and two, we can divide them by each other the south or A. So dividing equations two by equation one. OK. Then equation to T sine theta equals M sorry M prime A divided by tea cas theta it was M prime G. OK. Then we notice that we notice that TKTM prime Cancers M prime and sine theta divided by the cosine of theta equals the tangent of theta. So that means the tangent of theta equals A divided by G which means that the acceleration A is going to be equal to G multiplied by the tangent of theta. OK. Now that's a very important result because now we have an expression for our acceleration A let's call that equation four. Now the thing is we know what G is the acceleration due to gravity. That's uh that's the term that we're going to want in our final answer. But the problem is we don't know what theta is, we don't know what the angle theta is. So we'll need to figure theta out. Let's try to plug equation three. OK into equation two to see if that can help us any further because not this equation three has the equation two, sorry has theta in it. So plug in equation on three into equation two, then that means T sine theta where T equals ma is now going to become ma sine theta. So ma sine theta equals M prime multiplied by A. Since that's the case here, that means sine theta is going to be equal to M prime A divided by M and our ears cancer that tell us that sign theater. OK? I think I can make some space here actually because we could cancel, we could cancel a from our previous expression because a cancers A. So this tells us that sign theater equals M prime divided by M. Now how is this supposed to help us figure out the value of theta? What, what does it mean here? Well, if we think about this, OK. What is the sign of a of, of a, of an anger? Recall that the sine function is defined as the ratio of the opposite side to the hyperten. Also recall that the tangent is defined as the opposite side to the adjacent side. So if we can use the values of our opposite and hypotenuse for a triangle that we can form from our equation. OK, then we'll be able to get an expression in terms of M and M prime for the tangent of theta. OK. In other words, let's draw that out. OK. So solving for a 10 theater, what we're really saying is that the sine of theta can form a triangle with M from a right triangle. Sorry with our anger theta where M prime is opposite to theta. OK. And M is the hypotenuse of our triangle. OK. Recall that we know the tangent theater. Oops, let me get that right. We already know that the tangent of theta put it in blue here is equal to the opposite side. If I, if we label our triangle ABC, then it's gonna be the opposite side, which is A B ratio of the opposite side. AB two, the adjacent side BCE where we know that A B is M prime, all we need to do now is figure out the value of BC. While we're talking about a right triangle, we have two sides. And we know that by the Pythagorean theorem, let me make some space here by the Pythagorean theorem, the square of the longest, the square of the longest side. So that's M squared is going to be equal to the sum of the squares of the two shorter sides. So that's M prime M prime squared plus BC squared, which means if we can rearrange this formula BC squared is going to be equal to M squared minus M prime squared. Which if we square root both sides, that means BC is going to be equal to the square root of M squared minus M prime squared. So since we have that value for BC, now we can substitute it into our formula for the tangent of theta, which means the tangent of theta is M prime. OK. Divided by that's M prime divided by the square root of M squared minus M prime squared. No, this is a pretty important result because since we have a value for the tangent in terms of M and M prime, we can substitute it into our formula for A to get a value for a sorry, my apologies to get a value for our acceleration. A in terms of M and M prime. I'm just trying to draw the perfect rectangle here just to isolate that I think this works best. So now we can substitute our value for the tangent of theta into our formula for acceleration. So substituting substituting turn theater into equation four, the call equation four was G tan theta. So no, that means the acceleration is going to be equal to G multiplied by our value for tan theta which is M prime divided by the square root of M squared minus M prime squared. Know that we have a value for our acceleration in terms of mm prime. And G, that's what we need to put into our formula for our force F. So now let's replace. OK. So solving for F, solving for F no, that means our first F is going to be equal to our mass, which we said was the mass of the trolley, the mass of our ball M prime and the mass of our wine bottle M multiplied by this expression for a which is G multiplied by M prime divided by the square root of M squared minus M prime squared. Since that's the case, now let's combine our expressions here. So now we're going to get our force F to be equal to M plus M prime plus common M multiplied by M prime multiplied by G all divided by the square root of M squared minus M prime squared. Thus, this is the expression for the force that is acting on our trolley. If we go back to our answer choices, that means see is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

III) Determine a formula for the magnitude of the force F → exerted on the large block () in Fig. 4–56 so that the mass m_A does not move relative to m_C. Ignore all friction. Assume m_B does not make contact with m_C . <IMAGE>

Key Concepts

Newton's Second Law

Equilibrium Conditions

Force Interaction

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