Q76P - University Physics with Modern Physics

Question

Block A in Fig. P5.76 weighs 60.0 N. The coefficient of static friction between the block and the surface on which it rests is 0.25. The weight 12.0 N , and the system is in equilibrium. (a) Find the friction force exerted on block A. (b) Find the maximum weight w for which the system will remain in equilibrium.

                        

Step-by-Step Solution

Answer

 (a) The friction force exerted on the block A is 12 N .

(b) The maximum weight w is 15 N .

Step 1: Identification of the given data

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 $\mathrm{\mu }=0.25$ .
• The value of the weight is w = 12.0 N .
• The angle,  $\mathrm{\theta }=45°$

Step 2: Significance of the force

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.

Step 3: (a) Determination of the friction force

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

  $\mathrm{T}=\frac{\mathrm{w}}{\mathrm{sin\theta }}$                                                                                                                 ….. (1)

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

 

The equation of the friction force is expressed as:

 $\mathrm{F}=\mathrm{T} \mathrm{cos\theta }$

Here, F is the friction force.

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

 $$\begin{gathered} \mathrm{F}=\frac{\mathrm{w}}{\mathrm{sin\theta }}\mathrm{cos\theta } \\ =\mathrm{w} \mathrm{cot\theta } \end{gathered}$$

 

Substitute  $45°$for $\mathrm{\theta }$ 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. 

Step 4: (b) Determination of maximum weight

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

 $\mathrm{f}=\mathrm{\mu W}$

 

Here,   is the frictional force,   is the friction 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 .