(III) Three blocks on a frictionless horizontal surface are in contact with each other as shown in Fig. 4-54. A force $\vec{F}$ is applied to block A (mass $m_A$). ($a$) Draw a free-body diagram for each block. Determine ($b$) the acceleration of the system (in terms of $m_A, m_B,$and $m_C$ ), ($c$) the net force on each block, and ($d$) the force of contact that each block exerts on its neighbor. (e) If $m_A=m_B=m_C=$10.0 kg and $F=$96.0 N, give numerical answers to ($b$), ($c$), and ($d$). Explain how your answers make sense intuitively.

Answer

(a) In the free-body diagrams below, $\overline{\mathbf{F}}_{\mathrm{AB}}=$ force on block A exerted by block $\mathrm{B}, \overline{\mathbf{F}}_{\mathrm{BA}}=$ force on block $\mathrm{B}$ exerted by block $\mathrm{A}, \overrightarrow{\mathrm{F}}_{\mathrm{BC}}=$ force on block $\mathrm{B}$ exerted by block $\mathrm{C},$ and $\overrightarrow{\mathrm{F}}_{\mathrm{CB}}=$ force on
block C exerted by block B. The magnitudes of $\overrightarrow{\mathrm{F}}_{\mathrm{RA}}$ and $\overrightarrow{\mathrm{F}}_{\mathrm{AB}}$ are equal, and the magnitudes of
$\overrightarrow{\mathbf{F}}_{\mathrm{BC}}$ and $\overrightarrow{\mathbf{F}}_{\mathrm{CB}}$ are equal, by Newton's third law.
b)$a=\frac{F}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$
c)$F_{\text {A net }}=F \frac{m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$ $F_{\mathrm{B} \text { net }}=F \frac{m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$ $F_{\mathrm{Cnet}}=F \frac{m_{\mathrm{C}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$
d)$F \frac{m_{\mathrm{C}}}{m_{\Lambda}+m_{\mathrm{B}}+m_{\mathrm{C}}}$,$F_{\mathrm{AB}}=F \frac{m_{\mathrm{B}}+m_{\mathrm{C}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$,$F \frac{m_{\mathrm{B}}+m_{\mathrm{C}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$
e)$3.20 \mathrm{m} / \mathrm{s}^{2}$,$32.0 \mathrm{N}$,$64.0 \mathrm{N}$,$32.0 \mathrm{N}$

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Video Transcript

let's talk about this question. We are given that three blocks on a frictionless horizontal surface are in contact with each other. The fourth step is employed on Marseille. We have to draw a free body diagram of each block. So that's one. So let's do that. Uh Do we have friction? And we don't have any friction? So that's frictionless. So that's a good thing to hear. Let's talk about block. So for block A. We have the weight of, we have the reaction force of a uh due to the floor and we have this force which is acting on it and there there will be a reaction force. Let's call it R. One between block and block B. So this is far blockade. Likewise, if we draw for block B, we definitely have a weight of B. We have a reaction of uh we have actually the reaction for the upward reaction force is definitely gonna be seen for all. But okay, let's draw let's write it as a peak. So at BB has to uh to contact services each of the so uh able exert a reaction force on B. Which is our one over here and see will exert on B. Which is our to let's say over here. Likewise we have it for see where and we have W. C. We have the reaction foresee and we have just one force because the B. Will exert on C. Which is our too, so are too will act over here. So these are the free body diagrams of all the three blocks. The next part talks about. Let's see the acceleration of the system in terms of M E M B and M. C. So this is pretty much easy because the F net is equal to mass, which is the net mask um A plus M. B plus M. C. Times acceleration. Because we are taking this complete system as a unit that can be assumed as a as a blog which has an force acting uh F. And the mass having M A plus M. B plus M C. This is kids can be resumed like it. So the new mass for the system is I'm A plus mbps, M C. Which means that the acceleration is going to be forced over M. A plus M. B plus M C. So this would be the required acceleration. Uh Next talks about the net force which is acting on each block. The net force which is acting on each block. Okay. Ah Do we have to give in terms of M E M B M C. Or I think yes. So what we can do is the net force acting on each block. Mm. Ah We're here let's talk about block A uh So for the block the net force is definitely gonna be mass times acceleration because if we talk about net force on a let's call it a day. So that's gonna be mass of eight times acceleration. That's what we can do because they are moving on the same acceleration, right? And we already have the value of A. So their net force is going to be M. A. F over me plus MB Plus EMC likewise FB is gonna be same as MB times. So that's gonna be M. B. F over M plus M B plus M. C. And likewise Fc is going to be equal to M. C. Times, which is M. C. F over M. A plus M B plus M C. So these are the net forces on blocks. A. B. N. C. Talk about the next part. It has various parts uh for the force of contact on each block, the D block exerts on its neighbors. We need to find find the value of our one and artists. That would be pretty much easy. No, because we already got the net force. So let's talk about block for blockade. The net forces f minus R. One and that's equal to F. A. Right? That's equal to F. A. And a face already. We found that as a secret um A F for M M plus M B plus M C. Uh If he rearranged this, then the value of R. One S. F minus M. F. Over M. A. Plus and B plus M. C. And if we simplify this wedding in the common denominator, we get F times M. B plus M C over M. A plus M. B plus M C. So this is the reaction force one. Likewise. Uh Let's do the same kind of stuff on block B. So are one minus R. Two is something which we have to write our one minus R. Two is F. B. Which is M B. Mbf over M plus M. B plus M. C. Now we already have the value of our one. So the value of our two is going to be our one minus mbf over M plus M. B. Plus and illustrated as submission of them. So our one is now F. Times M. B plus M. C. Over the submission of em minus M. B. F over submission of them. So if you're gonna subtract, there's the denominators anyway, common. So that's going to be equal to M. C. Uh Times it's going to be M. C. Times F over M plus M B plus M C. That's going to be the value of the reaction force, are you? And the last part talks about if all the masters are equal and that's equal to 10 kg. The forces given, we have to give the numerical answer for part B. C. And D. So all the masses are 10. So let's talk about let's talk about part B first. So we're going to write part E. So we are going to find the acceleration now. The forces 96. And the forces 96. And the mods are 10 Kg. Hitch. So the acceleration is f over the sum of all the month. So that's going to be 96 over 30. Which is gonna be 96 or 30. There's going to be 3.2, 3.2. This is the value of uh and then we have to find uh then we have to find the reaction forces so that that is the acceleration times individual masses, so that's going to be seen for all. Uh so F net on is going to be 32 this is the same as definite of, be an F net of C. And then we need to find the value of R. one and R. two. So for our one, That's gonna be acceleration times and people SMC, it means that 3.2 times 20, so that's going to be 64. And the value of our two is gonna be uh just 32 because as for the calculation is coming as 32 new terms. So these are the final numerical answers. Thank you.

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