A hammer is hanging by a light rope from the ceiling of a bus. The ceiling is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is $\mathbf{56}\mathbf{.}\mathbf{0}\mathbf{°}$. What is the acceleration of the bus?

The given data can be listed below as:

The net force exerted by an object is equal to the product of the mass and the acceleration of that object. The net force is also independent on the velocity of that object.

The free body diagram of the hammer has been drawn below:

In the above diagram, the weight of the hammer mg is acting in the downwards direction and the tension T created by the hammer is acting in the direction making an angle of $56°$from the X-axis. The x and y-component of the tension T are $T\cos 56°$and$T\sin 56°$ acting along +x and +y-axis respectively.

From the above diagram, the net force acting in along x-axis is given as:

$$\begin{gathered} T\sin \theta =mg \\ m=\frac{T\sin \theta }{g} \end{gathered}$$ …(i)

Here, is the tension of the hammer, is the angle between the bus ceiling and the rope, is the mass of the hammer and is the acceleration due to gravity.

From the above diagram, the net force acting in along y-axis is given as:

$$\begin{gathered} T\cos \theta =ma \\ m=\frac{T\cos \theta }{a} \end{gathered}$$ …(ii)

Here, is the acceleration of the bus.

comparing the equation (i) and (ii)

$$\begin{gathered} \frac{T\sin \theta }{g}=\frac{T\cos \theta }{a} \\ a=\frac{g}{\tan \theta } \end{gathered}$$

Substitute the given values in the above equation.

$$\begin{gathered} a=\frac{9.8m/s^2}{\tan 56.0°} \\ =\frac{9.8m/s^2}{1.48} \\ =6.62m/s^2 \end{gathered}$$

Thus, the acceleration of the bus is$6.62m/s^2$ .