Question

1) What is the horizontal force on block A due to block B?2) What is the net horizontal force on block B?3) What is the horizontal force on block B due to block C?

Answer 1

Given data:

Mass of each block;

`m=12\text{ kg}`

Acceleration;

`a=1.2\text{ m/s}^2`

The free-body diagram for A,

The free-body equation for A is given as,

`F-N_1=ma\ldots(1)`

The free-body diagram for B is given as,

The free-body equation for B is given as,

`\begin{gathered} F_B=N_1-N_2 \\ ma=N_1-N_2\ldots(2) \end{gathered}`

The free-body equation for C is given as,

The free-body equation for C is given as,

`\begin{gathered} F_c=ma \\ N_2=ma\ldots(3) \end{gathered}`

Equating equation (2) and (3),

`\begin{gathered} N_1-N_2=N_2_{} \\ N_1=2N_2 \\ N_1=2ma \end{gathered}`

Part (1),

The horizontal force on block A due to block B is given as,

`\begin{gathered} F_(AB)=N_1 \\ =2ma \end{gathered}`

Substituting all known values,

`\begin{gathered} F_(AB)=2*(12\text{ kg})*(1.2\text{ m/s}^2) \\ =28.8\text{ N} \end{gathered}`

Therefore, the net horizontal force on block A due to block B is 28.8 N.

Part (2)

The net horizontal force on block B is given as,

`\begin{gathered} F_B=N_1-N_2 \\ =2ma-ma \\ =ma \end{gathered}`

Substituting all known values,

`\begin{gathered} F_B=(12\text{ kg})*(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}`

Therefore, the net horizontal force on block B is 14.4 N.

Part (3)

The horizontal force on block B due to block C is given as,

`\begin{gathered} F_(BC)=N_2 \\ =ma \end{gathered}`

Substituting all known values,

`\begin{gathered} F_(BC)=(12\text{ kg})*(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}`

Therefore, the horizontal force on block B due to block C is 14.4 N.