Question

In Figure 5-36, let the mass of the block be $\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kg}$and the angle $\mathit{\theta }$ be${\mathbf{30}}^{\mathbf{°}}$. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the resulting acceleration of the block.

Short answer

1. Tension in the Cord is$42\mathrm{N}$.
2. Normal force acting on the block is$72\mathrm{N}$.
3. When the cord is cut, the magnitude of the resulting acceleration isrole="math" localid="1657002235976" $-4.9\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

The given data

• Mass of the block,$m=8.5\mathrm{kg}$.
• Angle of inclination of motion, $\theta ={30}^{°}$.

Understanding the concept of force and free body diagram

From Newton’s second law, the net force acting on the body is equal to the vector sum of all the forces and is the product of the mass and acceleration of the body. The free-body diagram helps us to understand the forces acting on the body and their directions.

Using the free body diagram of the block, we can determine the tension in the block and the normal force acting on it.

Using Newton’s 2^nd law, we can find the magnitude of the acceleration.

Formulae:

Force according to Newton’s second law,$F=ma$ (i)

Here, F is the force, mis the mass and ais the acceleration of the body.

(a) Calculation of tension in the cord

Since the body is in equilibrium, the net force in the x direction on the body is zero. Write the all the forces in the x direction and substitute their values to calculate the tension as follows.

$$\begin{gathered} \sum _{}{F}_{\times }=0 \\ T-mg\sin \theta =0 \\ (\because \mathrm{the}\mathrm{tension}\mathrm{in}\mathrm{the}\mathrm{cord}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{sine}\mathrm{component} \\ \mathrm{of}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{block},\mathrm{mg}) \\ T=mg\sin \theta \\ =8.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin {30}^{°} \\ =41.65\mathrm{N} \\ \approx 42\mathrm{N} \end{gathered}$$

So, the tension value is

(b) Calculation of the normal force on the block

Since the body is in equilibrium, the net force in the y direction on the body is zero. Write the all the forces in the y direction and substitute their values to calculate the normal force as follows.

$$\begin{gathered} \sum _{}{F}_y=0 \\ {F}_N-mg\cos \theta =0 \\ (\because Thenormalforceisthe \\ perpendicularforceonthebook) \\ {F}_N=mg\cos \theta \\ =8.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \cos 30° \\ =72.05\mathrm{N} \\ \approx 72\mathrm{N} \end{gathered}$$

Hence, the value of the normal force is$72\mathrm{N}$

(c) Calculation of magnitude of acceleration

As the cord is cut, there is no longer tension in the string i.e. T = 0 N Hence in this case,

The net force is given by equation (i). Use equation (i) and substitute the values of mass and net force to find the acceleration of the body.

$$\begin{gathered} F_{Net}=ma \\ -mg\sin \theta =ma(\because Fromfreebodydiagram) \\ a=-g\sin \theta \\ =-9.8\mathrm{m}/{\mathrm{s}}^2\times \sin \left({30}^{°}\right) \\ =-4.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Here the negative sign shows that the acceleration is acting down the plane and its value is$-4.9\mathrm{m}/{\mathrm{s}}^2$.