Finding the Weight and Center of Gravity of a Heavy Electric Motor

[MAPgirl23]

Two people carry a heavy electric motor by placing it on a light board of length 2.50 m. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 560 N.

a) What is the weight of the motor?
b) Where along the board is its center of gravity located? Express your answer as a distance measured from the end where the 400 force is applied.

** I know that each force times its distance to the center of mass gives the same product (the moments are equal) but don't know how to do this problem.

 

[MAPgirl23]

thanks anyways, got it

 

[thepatientmental]

a) To find the weight of the motor, we can use the formula W = Fnet, where W is the weight, Fnet is the net force acting on the motor, and g is the acceleration due to gravity (9.8 m/s^2). In this case, the net force is the sum of the two forces acting on the motor, which is 400 N + 560 N = 960 N. So, the weight of the motor is 960 N.

b) To find the center of gravity of the motor, we can use the formula x = (F1d1 + F2d2)/Fnet, where x is the distance from the end where the 400 N force is applied, F1 and F2 are the forces acting on the motor, and d1 and d2 are the distances from the end where the 400 N force is applied to the center of gravity. In this case, F1 = 400 N, F2 = 560 N, d1 = 0 m (since the 400 N force is applied at this end), and d2 = 2.50 m (since the light board has a length of 2.50 m). Plugging in these values, we get x = (400 N * 0 m + 560 N * 2.50 m)/960 N = 1.46 m. So, the center of gravity of the motor is located 1.46 m away from the end where the 400 N force is applied.