A man raises a joist (length = l) by pulling a rope attached to point B. The centre of mass is located at the centre of the beam and the system is at equilibrium.Variable l αβValue 5 40 25Units [m] [degrees] [degrees]Hints: The reaction does not necessarily act along the joist. This question can be solved without the use of simultaneous equations (Use 3 Force Bodies).Given the information in the table above, determine:1) the angle of the resultant reaction, θ, at point A if at that point the ledge acts as a rough surface.

To solve this problem, we need to consider the forces acting on the joist. The forces are the weight of the joist acting downwards at its center of mass, the tension in the rope pulling upwards at point B, and the reaction force at point A.

1. First, we need to find the weight of the joist. This can be done by multiplying the mass of the joist by the acceleration due to gravity. However, since we don't have the mass of the joist, we'll represent the weight as W.

2. Next, we consider the tension in the rope. The tension can be split into two components: one acting horizontally (Tcosβ) and one acting vertically (Tsinβ).

3. The reaction force at point A can also be split into two components: one acting horizontally (Rcosθ) and one acting vertically (Rsinθ).

Since the system is in equilibrium, the sum of the forces in the horizontal and vertical directions should be equal to zero.

For the horizontal direction, we have:

Tcosβ - Rcosθ = 0 ----(1)

For the vertical direction, we have:

Tsinβ + Rsinθ - W = 0 ----(2)

From equation (1), we can express Rcosθ as Tcosβ.

Substitute Rcosθ in equation (2) with Tcosβ, we get:

Tsinβ + Tcosβtanθ - W = 0

Rearrange the equation to find tanθ:

tanθ = (W - Tsinβ) / Tcosβ

To find the angle θ, we take the inverse tangent of the right side:

θ = atan[(W - Tsinβ) / Tcosβ]

Without the values of W and T, we can't calculate the exact value of θ. However, this is the general method to find the angle of the resultant reaction at point A.