Question

Block B in Fig. 6-31 weighs $\mathbf{711}\mathbf{}\mathbf{N}$.The coefficient of static friction between block and table is $\mathbf{0}\mathbf{.}\mathbf{25}$; angle $\mathit{\theta }$is ${\mathbf{30}}^{\mathbf{0}}$; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary?

Short answer

The maximum weight of block A for which the system will be stationary is $102\mathrm{N}$.

Step-by-step solution

Given

Weight of the block B,$\mathrm{W}=711\mathrm{N}$

Coefficient of static friction between block and table,${\mathrm{\mu }}_{\mathrm{s}}=0.25$

Angle,$\mathrm{\theta }={30}^0$

Determining the concept

To find the weight of the block A, use Newton's 2nd law for the stationary system. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum ma$

where, F is the force, m is mass and a is an acceleration.

Determining the free body diagram

Free Body Diagram of the system:

Determining the maximum weight of block A for which the system will be stationary

By using Newton’s 2nd law of motion along the vertical direction to the block B,

$$\begin{gathered} N-W=0 \\ N=W \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_s={\mu }_{s}N \\ ={\mu }_{s}W \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the block B,

$$\begin{gathered} T\cos \left(\theta \right)-f_s=0 \\ T\cos \left(\theta \right)={\mu }_{s}W \\ T=\frac{{\mu }_{s}W}{\cos \left(\theta \right)} \\ =\frac{(0.25)\left(711\right)}{\cos \left(30\right)} \\ =205N \end{gathered}$$

Using the Newton’s 2nd law along vertical direction to the block A,

$$\begin{gathered} T\sin \left(\mathrm{\theta }\right)-\mathrm{W}\text{'}=0 \\ \mathrm{W}\text{'}=\mathrm{Tsin}\left(\mathrm{\theta }\right) \\ =205\sin \left(30\right) \\ =102\mathrm{N} \end{gathered}$$

Thus, the maximum weight of block A for which the system will be stationary is $102\mathrm{N}$.