Calculating Force Due to Inertia of a Block

[Dupain]

Homework Statement

Hi guys, I have a very beginner question:

Lets say we have a block(147mm*241mm) of mass M=5.1kg resting on a frictionless surface.
Now if a force of F, 483N is applied to one corner(A) of the block, the latter moves in a horizontal plane and at this instant the block does not have angular velocity.
How do we calculate the x-component of the acceleration of the opposite corner(B)??

I know i have to find the moment of inertia which i did using : 5.1/12 *(147^2+241^2)... then what should i do??

Homework Equations

Equation for inertia: M/12(L^2+b^2)
F-f=Ma

f=force due to inertia

The Attempt at a Solution

I have only calculated the inertia and i do not know how to continue

 

[KataKoniK]

from there.

Dear student,

It seems like you are on the right track with calculating the moment of inertia. Now, to calculate the x-component of the acceleration of the opposite corner (B), we can use the equation F-f=Ma, where F is the applied force, f is the force due to inertia, M is the mass of the block, and a is the acceleration. In this case, we are interested in the x-component of the acceleration, so we can write it as ax.

First, we need to find the force due to inertia, f. This can be calculated using the moment of inertia you have already found, by multiplying it by the angular acceleration. However, since the block is not rotating, the angular acceleration is 0, so the force due to inertia is also 0.

Now, we can rewrite the equation as F=Ma, and solve for ax. We know the applied force (F) and the mass (M) of the block, so we can plug those values in and solve for ax. This will give us the x-component of the acceleration at corner B.

I hope this helps you continue with your solution. Keep up the good work and don't hesitate to ask for further clarification if needed.