# 2 block friction problem

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?

## 2 Answers

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.

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 $$F > \frac{(\mu_1(m_1 + m_2) g + \mu_2 m_2 g)}{\cos \theta + \mu_1 \sin \theta}.$$