Question 4.SP.1: A fixed crane has a mass of 1000 kg and is used to lift a 24......

STRATEGY: Draw a free-body diagram to show all of the forces acting on the crane, then use the equilibrium equations to calculate the values of the unknown forces.

MODELING:

Free-Body Diagram. By multiplying the masses of the crane and of the crate by g = 9.81 m/s², you obtain the corresponding weights—that is, 9810 N or 9.81 kN, and 23 500 N or 23.5 kN (Fig. 1). The reaction at pin A is a force of unknown direction; you can represent it by components A_{x}~and~A_{y}. The reaction at the rocker B is perpendicular to the rocker surface; thus, it is horizontal. Assume that A_{x}, A_{y}, and B act in the directions shown.

ANALYSIS:

Determination of B. The sum of the moments of all external forces about point A is zero. The equation for this sum contains neither A_{x}~nor~A_{y}, because the moments of A_{x}~and~A_{y} about A are zero. Multiplying the magnitude of each force by its perpendicular distance from A, you have

+↺\Sigma M_{A}=0\colon              +B(1.5 m) − (9.81 kN)(2 m) − (23.5 kN)(6 m) = 0

     B = + 107.1 kN                B = 107.1 kN → ◂

Because the result is positive, the reaction is directed as assumed.

Determination of A_{x}. Determine the magnitude of A_{x} by setting the sum of the horizontal components of all external forces to zero.

\underrightarrow{+}\Sigma F_{x}=0\colon      A_{x} + B = 0

 A_{x} + 107.1 kN = 0

        A_{x} = − 107.1 kN                A_{x}= 107.1 kN ← ◂

Because the result is negative, the sense of A_{x} is opposite to that assumed originally.

Determination of A_{y}. The sum of the vertical components must also equal zero. Therefore,

+↑ \Sigma F_{y}=0\colon           A_{y} − 9.81 kN − 23.5 kN = 0

          A_{y} = +33.3 kN                    A_{y} = 33.3 kN ↑ ◂

Adding the components A_{x}~and~A_{y} vectorially, you can find that the reaction at A is 112.2 kN ⦩17.3°.

REFLECT and THINK: You can check the values obtained for the reactions by recalling that the sum of the moments of all the external forces about any point must be zero. For example, considering point B (Fig. 2), you can show that

+↺\Sigma M_{B} = −(9.81 kN)(2 m) − (23.5 kN)(6 m ) + (107.1 kN)(1.5 m) = 0