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Consider a block of mass $m_2$ placed on a heavier block of mass $m_1$. $m_2$ is tied to a wall with string. A force $F$ is applied on $m_1$ at an angle $\theta$ to the horizontal (upwards). The friction coefficient between $m_1$ and ground is $\mu_1$ and that between $m_1$ and $m_2$ is $\mu_2$. The problem is to find minimum force to start moving $m_1$.

My doubt is in finding frictional force at top of $m_1$ due to $m_2$. I found frictional force (at top of $m_1$) as $f=\mu_2\cdot(m_2\cdot g+F\sin\theta)$. But in the book the answer does not have the $F\sin\theta$ term. Shouldn't it be considered because it increases the normal reaction from $m_1$ on $m_2$ as the force is acting at an angle?

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A free body diagram of block 2 will show only four external forces acting on it: (1) gravity acting vertically downward (2) the normal reaction force of block 1 acting vertically upward (3) the tension in the string acting horizontally (say to the left) and (4) the friction force of block 1 acting horizontally (say to the right). Assuming the string doesn’t break and the vertical component of the $F$ doesn’t exceed the weight of both blocks, block 2 doesn’t accelerate horizontally or vertically and the sums of the horizontal and vertical forces are zero. Therefore the normal reaction force of block 1 on block 2 is simply the weight of block 1.

The vertical component of $F$ only reduces the reaction force, and consequently the maximum static friction force, of the ground on block 1.

Hope this helps.

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From the mental picture you have drawn (very well) in your question, the block $1$ with mass $m_2$ sits atop of the friction surface (coefficient $\mu_2$) of the block $2$ with mass $m_1$, which itself lies on a friction surface (coefficient $\mu_1$). The force $\vec{F} = F \cos \theta \hat{i} + F \sin \theta \hat{k}$ is applied to the block with mass $m_1$, where $0 \leq \theta$ and the unit vectors $\hat{i}$ and $\hat{k}$ indicate the horizontal and vertically upwards directions respectively. Finally a rope (with tension say $T$) is attached to the block $2$ preventing it from moving in the direction of the force $(\vec{F} \cdot \hat{i}) \hat{i}$.

Let $N_1$ be the magnitude of the normal reaction force acted by the friction surface on block $1$ and $N_2$ be the magnitude of the normal reaction force acted on the block $2$ by th esurface of block $1$. From the free body diagram, the static equations of motion for the $\hat{k}$ direction will read, $$0 = N_1 - (N_2 + m_1 g - F \sin \theta),$$ $$0 = N_2 - m_2 g,$$ for the blocks $1$ and $2$ respectively, due to the fact that neither block accelerates in the vertical direction. The static equation of motion for the block $2$ in the $\hat{i}$ direction is $$0 = T - \mu_2 N_2,$$ since the block will not (presumably) accelerate horizontally, while the dynamic equation of motion for the block $1$ in $\hat{i}$ direction is $$m_1 a_{1 \hat{i}} = F \cos \theta - \mu_1 N_1 - \mu_2 N_2,$$ where $a_{1 \hat{i}}$ is the horizontal acceleration of the block $1$. These calculations can be directly by doing the algebra, to explicitly find the magnitude of the force $\vec{F}$ such that $a_{1 \hat{i}} > 0$. Indeed, the required constraint is given as $$F > \frac{(\mu_1(m_1 + m_2) g + \mu_2 m_2 g)}{\cos \theta + \mu_1 \sin \theta}.$$

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