Two people carry a heavy electric motor by placing it on a light board 2.60m lon

Question

Two people carry a heavy electric motor by placing it on a light board 2.60m long. One person lifts at one end with a force of 410N , and the other lifts the opposite end with a force of 570N .
A)What is the weight of the motor?
w= 790N

B)Where along the board is its center of gravity located?
x= 1.51 m from the end where the 410- force is applied

C)Suppose the board is not light but weighs 190 N, with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case?
w= 790 N

D)Where is its center of gravity located?
x= ???? m from the end where the 410- force is applied (I can't answer this part plz help)

Explanation / Answer

answer of the first case is 980 N .

answer of second case is 1.51 m

answer of third is 790 N

answer for the fourth is as follows

let the distance between the motor (COM) and plank (COM) =x

then distance of combined (COM) of the sytem from plank (COM) =

790*x(1)= 190*(x-x(1))

980*x(1)= 190*x

x= 5.157x(1)

now torque must be zero of the whole system

so 190*x+410*(1.3+x)= 570*(1.3-x) + 790*x/(5.175)

827.35*x=208

x=0.251

so we get that from the plank (COM) the distance of new (COM)=.251m

hence the distance of the new (COM) from the 410 N end = 1.551 m

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