Determine the internal shear force at point E.Determine the internal moment at point E.Determine the internal normal force at point F.Determine the internal shear force at point F.Determine the internal moment at point F. Consider the compound beam shown in (Figure 1).Suppose that P = 840 N , P2 = 1250 N ,w = 370 N/m , and point E is located just to the left of840 N force. Follow the sign convention.Figure1 of 1A|E BF1.5 m 1 m---2 m+1m-1.5 m 1.5 m-

Consider the member BC and draw the free-body diagram as follows. Take the sum of the moment about point B and equate it to zero. ∑MB=0Cy×3-125045×2=0Cy=666.67 N Take the sum of the moment about point C and equate it to zero. ∑MC=0-By×3+125045×1=0By=333.33 N Write the equation of the sum of horizontal forces. ∑Fx=0125035-Bx=0Bx=750 NConsider the segment BE and show the internal loading as follows. Write the equation of the sum of horizontal forces. ∑Fx=0750-NE=0NE=750 N Write the equation of the sum of vertical forces. ∑Fy=0VE-840-333.33=0VE=1173.33 N Therefore, the internal shear force at E is 1173.33 N. Take the sum of the moment about point E and equate it to zero. ∑ME=0-ME-333.33×1=0ME=-333.33 Nm Therefore, the internal moment at point E is -333.33 Nm.Consider the segment CF and show the internal loading as follows. Write the equation of the sum of horizontal forces. ∑Fx=0NF=0 Therefore, the internal normal force at point F is zero. Write the equation of the sum of vertical forces. ∑Fy=0-VF-666.67-370×1.5=0VF=-1221.67 N Therefore, the internal shear force at F is -1221.67 N. Take the sum of the moment about point F and equate it to zero. ∑MF=0MF+370×0.75×0.752+666.67×0.75=0MF=-604.065 Nm Therefore, the internal moment at point F is -604.065 Nm.