Given in the figure are two blocks $A$ and $B$ of weight $20\,N$ and $100\,N$, respectively. These are being pressed against a wall by a force $F$ as shown in figure. If the coefficient of friction between the blocks is $0.1$ and between block $B$ and the wall is $0.15$, the frictional force applied by the wall in block $B$ is
To solve this problem, we need to determine the frictional force exerted by the wall on block $B$. Given the weights of blocks $A$ and $B$, the coefficient of friction between block $B$ and the wall, and the force $F$ applied, we can use the following steps:
Calculate the normal force $N_B$ exerted by the wall on block $B$. This normal force is equal to the horizontal force $F$ applied to push blocks $A$ and $B$.
The maximum frictional force $f_{\text{max}}$ that can act on block $B$ is given by $f_{\text{max}} = \mu N_B$, where $\mu = 0.15$ is the given coefficient of friction.
Since block $B$ has a weight of $100\,N$, the frictional force has to balance this weight to prevent block $B$ from sliding down. Therefore, $f_B = 100\,N$.
Set the equation of equilibrium for block $B$:
$$f_{\text{friction}} = f_B + 20\,N \times 0.1$$
Substitute the known values:
$$f_{\text{friction}} = 100\,N + 2\,N = 102\,N$$
To balance both blocks, calculate the required force to achieve the frictional force of $120\,N$. Therefore:
$$f_{\text{friction}} = \mu N_B$$ leads to
$$120\,N = 0.15 \times F$$
Solve for $F$:
$$F = \frac{120\,N}{0.15} = 800\,N$$
Thus, the frictional force applied by the wall on block $B$ is indeed $120\,N$.
