enter image description here

So , I was trying to solve it. Then , I notice that acc of 2kg mass has 2effects. One acc from the force to left and one from the Question , it says acceleration to be up. Then , the 2kg mass would go diagonally and not straight .

I am not getting how to solve this Question further . Is it possible to solve this Question from inertial frame.

  • 1
    $\begingroup$ Can you edit your question title to be a bit more explicit about what 'Q' is about? That way you might attract people who understand the topic. "Compound accelerations" might be appropriate to include. $\endgroup$
    – Transistor
    Commented Feb 23, 2021 at 16:19
  • $\begingroup$ is there friction in the problem? $\endgroup$
    – NMech
    Commented Feb 23, 2021 at 16:43
  • $\begingroup$ No friction .@NMech $\endgroup$
    – Rider
    Commented Feb 23, 2021 at 16:46
  • $\begingroup$ @NMech Yes , I have sir. $\endgroup$
    – Rider
    Commented Mar 11, 2021 at 17:20

3 Answers 3


Assuming there is no friction.

First of all lets define the coordinate systems. In the image there is the inertal system (lets denote it Oxyz) and there is the parallel axis accelerating system with the 10kg mass (let's denote it as P - from the pulley). In general, the following equations hold:

$$x_B = x_A + x_{B|A}, \qquad \ddot{x}_B = \ddot{x}_A + \ddot{x}_{B|A}$$

mass 2

On mass 2 you will have only 3 forces. The weight $m_2 g$, the tension of the rope $T$ and the reaction $N$ from the contact between mass 2 and mass 10. the following equation can be written:

$$m_2 \ddot{x}_{2|O} = N $$ $$m_2 \ddot{y}_{2|O} = T-m_2g $$


  • $\ddot{x}_{2|O}$ the horizontal acceleration on mass 2 wrt the inertial system O. It is worth noting that $\ddot{x}_{2|O} = \ddot{x}_{10|O}$ ,because mass 2 moves horizontally attached to mass 10 (when they are moving in the same direction).
  • $\ddot{y}_{2|O}=4 [m/s2]$ the vertical acceleration on mass 2 wrt the inertial system O

therefore the above two equations can be rewritten (assuming gravity is g=10m/s2) as:

$$m_{2} \ddot{x}_{10|O} = N $$ $$T = m_2 (g+ \ddot{y}_{2|O}) =28 N$$

mass 3

On mass 3 only the wire force T is applied. In the system P ( which is moving with the pulley) we can see that: $$ \ddot{x}_{3|P} =\ddot{y}_{2|P} = 4[m/s^2] $$

Therefore, the acceleration of mass 3 wrt to O would be: $$ \ddot{x}_{3|O} = \ddot{x}_{P|O} + \ddot{x}_{3|P} $$

where: $\ddot{x}_{P|O} = \ddot{x}_{10|O} = \ddot{x}_{10} $

So on the inertial system the following equation holds $$ m_3 \ddot{x}_{3|0} = -T $$ $$ m_3 (\ddot{x}_{3|P} + \ddot{x}_{10|O}) = -T $$ $$ m_3 (4[m/s^2] + \ddot{x}_{10} ) = -T $$ $$ \ddot{x}_{10} = -4[m/s^2] - \frac{T}{m_3} $$ $$ \ddot{x}_{10} = -\frac{40}{3}[m/s^2]$$

mass 10

On the large mass the following equation describes the motion:

$$m_{10} \ddot{x}_{10} = F -N-T$$

That can be rearranged to:

$$ F = m_{10} \ddot{x}_{10} +N + T = m_{10} \ddot{x}_{10} +m_{2} \ddot{x}_{10|O} + T= m_{10} \ddot{x}_{10} +m_{2} \ddot{x}_{10} +T $$

$$ F = (m_{10} +m_2) \ddot{x}_{10} + T=188 [N] $$

  • $\begingroup$ There are only two moving parts in this problem - 2kg block and 10 kg block, that produce work. The 3kg block stays still, its position changes only due to the relative motion of the 10kg block, thus does no work. With respect to the axis of the pulley, assume the 2kg block moves a distance x upward towards the pulley, the cable must be lengthen a distance x horizontally away from the pulley, as well as the 10kg block but moving in te opposite direction. So the equation of works can be expressed as m_2gx+m_2ga = m_10gx, or m_2(g+a)x = Fx, Thus F=2*(9.81+4)=27.62N $\endgroup$
    – r13
    Commented Feb 25, 2021 at 0:23

Assuming no friction, the force on the cable is:

$$ F_c= 2g+ (2+3)4=39.6kg= (3-2)*\alpha$$


F must accelerate the block and counter the tension in the cable.

$$F= 39.6*10kg+39.6=435.6kg$$

  • $\begingroup$ Could you please explain the values you have written sir please? What are variables associated with them. $\endgroup$
    – Rider
    Commented Feb 23, 2021 at 16:48
  • $\begingroup$ Also , didn’t you need to non inertial frame. $\endgroup$
    – Rider
    Commented Feb 23, 2021 at 17:00
  • $\begingroup$ I am on the taxiway, ready for take off. will respond later. thanks. $\endgroup$
    – kamran
    Commented Feb 23, 2021 at 17:11
  • $\begingroup$ Sure. Enjoy flying . $\endgroup$
    – Rider
    Commented Feb 23, 2021 at 17:33

This is a simple static problem, with the assumption that all contact surfaces are frictionless. In the static position (no motion), the cable has a tension equal to 2g. Then, in order to move the 2kg mass upward, there is an additional tension that equals to 2a, so at motion under given acceleration, the total tension in the cable will be massx(g+a), that is T = 2x(9.81+4) = 27.62N. Since the tension is constant throughout the cable, the force on the 10kg block must equal to -27.62N, the minus sign indicates the push force is in direction opposite to the tension.

A final check needs to be made to ensure the 3kg mass is adequate to resist the tension without displacement, that is easily proved, since the weight of the mass 3x9.81 = 29.43N > 27.62N, check!

  • $\begingroup$ IMHO, this is not a statics problem. If the cable tension is equal to $2g$ the sum of forces on the mass would be zero, therefore the mass would not accelerate. Therefore you would not achieve the 4m/s2 acceleration. $\endgroup$
    – NMech
    Commented Feb 24, 2021 at 6:40
  • $\begingroup$ @NMech, This is a well designed homework problem. It tricks people to think this is a kinetic problem, but the solution only involves/requires force equilibrium in two axes. $\endgroup$
    – r13
    Commented Feb 24, 2021 at 16:36
  • $\begingroup$ If I understand correctly you are saying that the mass of the 10 kg does not affect in anyway the magnitude of the force? $\endgroup$
    – NMech
    Commented Feb 24, 2021 at 16:50
  • $\begingroup$ @NMech, The 10kg block needs to move to produce the counter force. At first impression, it involves motion - a none static event, however, result of the motion F = ma can be consider (in my opinion) static in nature, as other parameters of motion (t, s, v) are not required in solving this problem. $\endgroup$
    – r13
    Commented Feb 24, 2021 at 17:52
  • 1
    $\begingroup$ I think the only think we can agree is that we are disagreeing. :-) Take care. $\endgroup$
    – NMech
    Commented Feb 25, 2021 at 17:46

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