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Would anyone be able to help with this Newtons law problem
Block A rests on top of block B as shown in (Figure 1). The table is frictionless but there is friction (a horizontal force) between blocks A and B. Block B has mass 6.00 kg and block A has mass 2.00 kg. If the horizontal pull applied to...
m1 = 4 kg, m2 = 12 kg, m3 = 8 kg. k = 327 N/m for all three blocks. The elevator accelerates upwards at 3.8 m/s^2.
Net force of block one would be equal to force applied by top spring minus weight of system, since top spring is holding all 3 blocks.
F1 = 4*3.8= Fs,top  Wsystem = Fs,top ...
In both the cases 7 kg mass accelerates towards the right because of the 50N force. The unbalanced forces in both the cases are the force of gravity due to 5kg block and force of friction. Applying Newton's second law of motion to cases 1 and 2 yields the following results for acceleration...
I have calculated the acceleration of truck B from v=u+at as 5ms/s. The force applied to truck B is therefore 5x10=50N.
I am unsure whether this question is poorly worded, but I feel a reasonable assumption is that the force applied to truck A would be the same as truck B, without knowing its...
Assume that the lengths of various sections of the string at an instant are as shown. So we can say that
##l_1 + l_2+l_3= k## at an instant of time ##t##, where ##k## is a constant.
Taking the second derivative of both sides w.r.t. ##t##, we get $$\frac {d^2{l_1}} {dt^2} +\frac {d^2{l_2}} {dt^2}...
I don't know what to do as I know the forces acting on it but not the scalar like I just don't understand the question with two different pulleys I know the tension same everywhere
This problem is similar to what I have done before here. I think since the system is in equilibrium, that is both bodies are at rest, net force on each should be zero. So to balance the forces in all directions we need only friction forces on each in upward direction. So the force on B due to A...
Since no body accelerates so net force is zero. Force on each mass is zero. T1 and T2 both are 60N.
Edit: since there is a force applied so there is acceleration on friction less surface.
1. ##f_k\cos\thetaT\cos\theta+F_n\cos\alpha=m_2a_x##
2. ##f_k\sin\theta+T\sin\theta+F_n\sin\alpham_2g=m_2a_y##
3. ##Tm_1g=m_1a_y##
I am unable to get anywhere. There are accelerations in x , y directions.
I need the value of acceleration. Then I can simply use ##s=ut+\frac12at^2##
I was researching a problem that had once been posted here by someone else, and that had subsequently been posted (couched in somewhat different terms) here by me, and in doing the research, I ran across the following, which I think is the wellstated 'original version' of the problem.
The...
$$a= 40/(10+6+4)$$
$$a=2 m/s^2$$
Taking one mass of 10 kg.
$$T40=10(2)$$
$$T=20 N$$
This is correct.
But if I make the eqn of the system then
$$40+TT+TT=20(2)$$
I have also drawn the diagram. It looks like the second body m2 is subject to no force. But it’s accelerating. How?
Hello, everyone.
This problem is easy if you assume that the friction between the blocks and the platform is the maximum possible. Then the normal force is 0. But how can you show that the normal is 0 even when the friction is less than maximum?
Thank you for your help.
hello , I am adding pdf with the questions and what i tried to do to solve them.
in the first question my teacher just taught us that if there is a frictionless floor and i push the upper block (m) with force F toward me then if the static friction will be toward me too. so the lower block M is...
Hello, thanks for the attention. Well, knowing that the only acting force is the gravitational force, I stated that ##U=MgH## for the ##M## mass block and that ##U=mgh## for the ##m## mass block. After that I divide the two and got the relationship for the alternative "a". For alternative "b" I...
Let us assume that there are 5 blocks , each of weight W. The blocks are held horizontally by applying a high enough force, ( say N ) to the leftmost and rightmost blocks.
Let us say the blocks are numbered 1 to 5 from left to right.
So, if i try to draw the free body diagrams, then block 3 will...
I know string theory is already controversial enough and may even be falling out of favor among many physicists. But assuming it is right, I have heard string theorists say that strings are the "fundamental building block" of the universe. How do they make this assertion so confidently...
From Kleppner's Intro to Mechanics (Example 4.7, wording not exact): Two identical blocks a and b each of mass m slide without friction on a straight track. They are attached by a spring with unstretched length l and spring constant k; the mass of the spring is negligible compared to the mass of...
A force P is exerted on two articulated rods that are connected to two small blocks A and B. Both blocks have the same weight G. The magnitude of the force P = 1.26 G. The static friction coefficient between the blocks and the ground is 0.3. The mass of the rods may be neglected.
It can be...
the acceleration of the center of mass is ##a_{cm} = F/(M+m)##
I considered the forces on the block of mass m(when the system is at maximum extension) I got the equation $$kx  \frac {mF}{(M+m)} = ma_{cm}$$
and from that I got the value of the maximum extension $$x = \frac {2mF} {k(M+m)}$$ which...
if the applied force F >(M+m)g/##\mu## the force of friction between the two blocks should become f=N##\mu## and since the normal reaction will be greater than what it should be to keep the block m at rest with respect to M, will the mass m accelerate upwards.
in short,for let's say a very...
So, for this question first I did a free body diagram in the perpendicular xy axes, and, I got some equations with the normals, but the answer is independent of normal forces. So, I'm not sure how to eliminate the two normals. Further I find it quite weird that big blockA doesn't come into the...
"The displacements of the blocks
from equilibrium are both measured to the right. Block 1 has a mass of 15 grams and block 2 a mass of 10 grams. The spring constants of the springs are shown in dynes/cm."
I don't know if i understood very well the notation, but i interpreted as F(t) acting only...
I've problems understanding why the kinetic energy of the string is only
$$T_{string}=\frac{1}{2}m\dot{y} $$
Why the contribution of the string in the horizontal line isn't considered?
Let:
Smaller block = m1 = 1 kg
Large block = m2 = 2 kg
Coefficient of friction between the two blocks = μ1 = 0.2
Coefficient of friction between larger block and floor = μ2 = 0.3
Tension connecting two blocks through two pulleys = T
Angle between tension and horizontal = θ = 37o
Friction between...
So far I found the answer for a and b, but when I attempted to do the other ones I was completely lost.
A.) P= MV
M = 25g = .025kg
V = 18
.025 * 18 = .45kg*m/s
B.) KE= 1/2 mv^2
1/2 (.025)(18)^2
4.05 J
Hi,
I posted a similar question recently and gained some insight on these types of problems. However, I am slightly stumped on how to approach this variation of the problem.
So I know that:
 there is no net change in enthalpy of the blocks and the engine as the processes are reversible
...
I know that I am supposed to use the equation I wrote, but how do I find the maximum force of static friction and coeffcient of static friction if none of them are given beforehand? I can't have to unknowns in an equation. We then did the same thing, the only difference was then to measure the...
Hi,
I am quite confused about how to approach this problem. I have seen variations of this problem where there is a heat engine between two blocks, but in this case the surroundings are massless, so I don't believe that approach will work here.
Method:
I have first started with the case that...
I am thinking about solving it this way.
The first three equation is from Newton law.
Then the forth one is the constrain equation, after that I simplified the first three equations, and I am thinking about plugging them into the forth equation and solve for T after that I get the value of...
Free body diagram one each object:
Block 1:
Normal force upwards, weight downwards, tension to the right, friction to the left
Block 2:
Normal force to the right, tension upwards, weight downwards, friction upwards
Cart:
Contact force with block 1 downwards, contact force with block 2 to the...
I honestly have no idea why I'm getting this question wrong, because it seems fairly straightforward.
I thought that treating the two blocks as one object would work, so 5 + 3 = 8. With Fg = Fs, ma = kx. Then, 8 x 9.8/1300 = x. x would be 0.06. That's been marked as an incorrect answer, though...
Proof: Let ##B = \lbrace a \rbrace \subseteq A## and ##\rho \in S_4##. We have two cases, ##\rho(a) = a## in which case ##\rho(B) = B##, or ##\rho(a) \neq a## in which case ##\rho(B) \cap B = \emptyset##. Its clear that ##\rho(A) = A##. So these sets are indeed blocks. Now let ##C## be any...
I wrote the equations for ##m_2## with respect to the room, which are:
##x) TFr_1f_1 *=0##
##y) N_1P_1=0##
For ##m_1## we hace:
##x) N_2=f_2 *##
##y) T+Fr_2 P_2=0##
Where ##f*_1## and ##f*_2## are the pseudoforces and ##Fr_1## and ##Fr_2## are the friction forces that contain ##\mu##...
The figure is shown above. In order to facilitate solution, I need to separate each block and draw the freebody diagram (FBD) with all the forces acting on it.
1. Let me start with the lowest block. Let me put its FBD as shown below.
For forces in the vertical direction, ##N_3 = N_2 + m_3 g...
Since they're all in the same liquid I'm assuming the buoyant forces would be the same on each block. But then I think about the volumes of the blocks, and them being different. I'm not sure if the block's volume would affect the buoyant force. Any help would be great, thanks!
The red dots show the CM of each block. ##x## is the amount by which the upper block overhangs the lower block. The blue (dashed) line shows the CM of the combination. For maximum overhanging, this line lies on the edge of the table below. By symmetry, the CM lies exactly midway between the two...
I saw this general formula:
##w_{0} = \sqrt{\frac{k}{m}}##
In my case both masses after collision create connected system, so ##w_{0} = \sqrt{\frac{k}{m+M}}##
Plugging it into ##\omega = \sqrt{\omega_{0}^2  \beta^2}## gives :
##\omega = \sqrt{\frac{k}{m+M}  \beta^2} = \sqrt{80  21^2} <...
Summary: In the situation given, all the surfaces are friction less, pulley is ideal and string is light, F=mg/2, Find the acceleration of block 2 ?
Here's the diagram
My attempt (i) I tried to solve the question by making FBD of the two blocks, but i am not able to draw all the forces...
Equation 1: T = Ma1
Equation 2: mg  2T = ma2
Equation 3: a1 = 2a2
Since a1 = 2a for equation one I get T = 2Ma1
mg  4Ma1 = ma1
mg = 4Ma1 + ma1
mg = 4a1(M+m)
a1= mg / 4M+m
Not sure if this is correct. Can someone please help to make sure I'm doing this right?
Thanx
Homework Statement
hey I need help with this question I have this sketch and the tell me that m1= 2 kg, m2= 5 kg the say that we don't have friction between m2 and the floor the kinetic friction between m1 and m2 is 0.2 and the static friction is 0.5 and we have the f  f pushing m1
Homework...
Homework Statement
[/B]
Use MATLAB to find the current flowing through 2Ω resistance
Homework Equations

The Attempt at a Solution
Here's my work :
The DC voltage sources won't connect to the resistances , I get a red dotted line.
Where am I doing wrong ?
Your help is appreciated .
Homework Statement
Two blocks of iron, one of mass m at 10.0C and the other of mass 2m at 25.0c, are placed in contact with each other. If no heat is exchanged with the surroundings, which of the following is the final temperature of the two blocks?
A)10
B)15 .
D) 20C ( this is the answer)
The...