Answer to Question #108238 in Mechanics | Relativity for let me know

Question

[img]https://upload.cc/i1/2020/04/06/zhY4Dw.jpg[/img]



question is in picture, d8=4

Answer 1

Explanation:

• Without any externally applied force (like F in the figure) this system cannot be held in against the wall as no any resultant force could be generated, hence no any static frictional force which could balance the weight of the 2 blocks.
• What happens here is, force F is applied along a diagonal direction whose horizontal component provides force against the wall while the vertical component provides force to balance or move the system upwards.
• B
• Being in balance or moving upwards depends on the nature of the application of the force : F.
• During the application of equations it is convenient to consider both blocks as a one system instead of considering them individually.

Notations:

• Please refer to the free body diagram attached.
• R= Resultant force generated at the surfaces in contact of the 2 blocks.
• S= Resultant force generated between the wall & the block B (Only the component on the block B is shown in the free body diagram)
• f_d = dynamic frictional force generated on block B (static frictional force is = f_s where needed)
• 2g & 5g = Weights of the both blocks

Calculations:

1).Free body diagram

2).Applying F=m*a on B towards the wall,

$\begin{aligned} F*cos(40)-S& = m*0\\ S &= (180+2*4)*cos(40)\\ S &= 144.02N \end{aligned}$

Now, maximum static frictional force that could generate,

$\begin {aligned} f_{s}&= 0.3*S\\ & = 43.21N \end{aligned}$

Vertical component of the force F,

$\qquad= F*sin(40)\\ \qquad= 120.84N$

Total weight of the system

$\qquad = 2g + 5g\\ \qquad = 70N$

Now, it seems that the applied force may lift the blocks upward as the vertical component is larger than the weight of the blocks.

In such a phenomenon, the wall tries to stop that movement by generating a frictional force downwards.

Let's check whether this movement could be stopped by the maximum static fictional force available.Then the total downward force is,

$\qquad = 70N + 43.21N\\ \qquad = 113.21N$

It seems that the upward force is greater than the total downward force, making it is possible for the blocks to move upwards. So the blocks would move after applying this force.

3).So it moves upwards. An unbalanced force is generated upwards. Therefore, an acceleration is generated (F=m*a) & the direction of the acceleration is upwards.

Soon after it starts moving, the frictional force reduces down to the dynamic frictional force.

Therefore,

Dynamic frictional force applied to the block ,

$\qquad = 0.2*S\\ \qquad = 28.80N$

Therefore, the new downward force is,

$\qquad = 70N + 28.80N\\ \qquad = 98.80N$

Now, apply F=m*a upwards the system.

$\begin{aligned} 120.84 - 98.80 &= (2+5)*a \\ 22.04 &=7a \\ a &= 3.15 ms^{-2} \end{aligned}$

Acceleration = 3.15 ms^-2 (Upwards)

Hope you understand!