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I draw one for you,hope it helps!

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Although you do not need it you have miscalculated the position of the centre of mass of the arrangement of squares. About corner $O$ the two left-hand squares produce a anticlockwise torque whilst the right-hand square, the normal reaction of the squares due to the ground $N$ and the force $F$ all produce a clockwise torque about corner $O$. Without the ...

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Due to the fact that the body in equilibrium, all forces must cancel each other. Which forces must cancel each other when a body is in equilibrium? When a body is in equilibrium, resultant of forces acting on it must be zero. In current question, we have three bodies those are in equilibrium (man, pen and the earth). So, net force exerted on each of them ...

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The object will topple about the lower RH corner. Take moments about this corner. The net clockwise moment must be > 0 for the object to topple. There is no need to find the CM. You can calculate moments for each cubical component of the object.

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You claim the horizontal distance of the CoG ($X$) from the bottom left corner is $\frac{5a}{12}$ (I've not verified this). That is the distance $|OP'|$ in my version of the diagram. The distance $|PP'|$ is therefore: $$|PP'|=\frac{a}{2}-\frac{5a}{12}=\frac{a}{12}$$ When you start exerting the force $F$, a torque about the point P, which is $F\times \frac{... 0 The body will start to rotate when the torque is barely above zero. As you apply the force, the normal force made by the floor will move to the right to prevent rotation, until it reaches the right corner of the bottom square. After that it cannot keep moving. Thus to solve the problem make force diagram to get$N$in terms of$F$, and impose a torque equal ... 1 A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states for whatever system you are considering. So, for your system to experience a reversible process, its pressure and temperature must differ only slightly from that of its surroundings throughout the entire process. And there can be no spatial temperature or ... 1 It depends on what you consider to be the system. If the system is the entire container, then there are no thermodynamic operations, quasistatic or not, on the system by the external environment. And as you said, the system is not in thermal equilibrium. If you talk about a thermodynamic operations you need to define a system and an environment, in this ... 0 First, I should say that$\Delta W$and$\Delta Q$don't have meaning.$\Delta$exists for state functions, but work and heat are path functions. A system doesn't have heat or work. Heat and work are recognized when they are transmitted through system's boundary. Second, the correct equation is$\Delta U=mC\Delta T$. So, I assume that you want to determine$...

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The speed is constant but the velocity is not constant. These stars are constantly changing direction, just like a satellite in orbit. Because of this change in direction as the stars orbit there is a change in velocity and thus there is an acceleration, meaning there is a net force. As for why that specific force, set your focus on one star and fill in ...

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The answer is no. The total torque on the system given in terms of the center of mass coordinates is $$\vec\tau=\vec R_{cm}\times\vec F^{ext}+\sum_i\vec r_i'\times\vec F_i^{ext}$$ where $M$, $\vec R_{cm}$ and $\vec F^{ext}$ are respectively the mass of the system, the center of mass position vector and the total external force on the system. The vector $\... 2 Free body diagram of C is as below: Am I missing something important in interpreting the free body diagrams? Yes. You have assumed that$\vec N+\vec F+\large{\frac 12 m_C\vec g}=\vec 0$. But, this is not correct. The correct equation is: $$(\vec N)_y+(\vec F)_y-\large{\frac 12}m_Cg=0$$ or$\$N\sin(\large{\frac \alpha 2})+F\cos(\large{\frac \alpha 2})=\...

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