An athlete at the gym holds a 3.0 kg steel ball in his hand. his arm is 78 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder. what is the magnitude of the torque about his shoulder due to the ball and the weight of his arm if he holds his arm.
a. straight out to his side, parallel to the floor?
b. straight, but 45° below horizontal?

a. 34.6 Nm

b.24.4 Nm

Explanation:

a.

78 cm = 0.78 m

W = F =mg

m1 = mass of steel ball = 3 kg

m2 = mass of long arm = 3.8 kg

moment due to steel ball = Fd =(m1*g)*(0.78)= (3*9.81)(0.78)=22.95 = 23 Nm

moment due to arm =Fd=(m2*g)*(0.78*0.4)= (3.8*9.81)(0.312)=11.63 = 11.6 Nm

net moment = 23 +11.6 = 34.6 Nm

b. now in this the angle will change the perpendicular moment arm

   moment due to steel ball = (3*9.81)*(0.78cos45) = 16.23 =16.2 Nm

   moment due to arm = (3.8*9.81)(0.4*0.78cos45) = 8.22 = 8.2 Nm

net moment = 16.2 +8.2 = 24.4Nm

the answer you are looking for is d. there are no holes for blood to enter or leave the system. your welcome

explanation:

to push or pull an object in a certain direction is known as force. a force in moving an object from one place to another.

force depends on the mass of an object. so, more is the mass of an object more will be the force required to move the object.

thus, we can conclude that the use of force to move an object is push or pull.