91Ó°ÊÓ

The given data can be listed below as:

• The weight of the block A is W = 60.0 N .
• The friction coefficient between the surface and the block is $\mathrm{μ}=0.25$.
• The value of the weight is w = 12.0 N .
• The angle, $\mathrm{θ}=45°$

The force is described as the influence that contributes to the motion of an object. The force of an object is directly proportional to the mass and the acceleration of that object.

According to the diagram, the equation of the tension of the wire can be expressed as:

$\mathrm{T}=\frac{\mathrm{w}}{\mathrm{²õ¾\pm ²Ôθ}}$ ….. (1)

Here, T is the tension of the wire, w is the weight and$\mathrm{θ}$is the angle subtended by the weight and the block.

The equation of the friction force is expressed as:

$\mathrm{F}=\mathrm{T}\mathrm{³¦´Ç²õθ}$

Here, F is the friction force.

Substitute the value of the equation (1) in the above equation.

$$\begin{gathered} \mathrm{F}=\frac{\mathrm{w}}{\mathrm{²õ¾\pm ²Ôθ}}\mathrm{³¦´Ç²õθ} \\ =\mathrm{w}\mathrm{³¦´Ç³Ùθ} \end{gathered}$$

Substitute $45°$for $\mathrm{θ}$and 12 N for w in the above equation.

$$\begin{gathered} \mathrm{F}=\left(12\mathrm{N}\right)\cot 45° \\ =12\mathrm{N} \end{gathered}$$

Thus, the friction force exerted on the block A is 12 N.

The equation of the frictional force on the weight is expressed as:

$\mathrm{f}=\mathrm{μ°Â}$

Here, is the frictional force, is thefriction coefficient between the surface and the block and is the weight of the block A.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{f}=\left(60\mathrm{N}\right)\left(0.25\right) \\ =15\mathrm{N} \end{gathered}$$

In order to remain equilibrium, the frictional force should be equal to the block’s weight.

Thus, the maximum weight w is 15 N .