A man is attempting to raise a 7.5-m-long,

or approximately to two significant figures

Explanation:

Question (clarified by question poster): "With what force does the man have to pull on the rope to hold the pole motionless in this position?"

A FBD helps with identifying the forces on the pole. Call the axis parallel to the pole the x-axis. Directly perpendicular to this is the y-axis.

The force of gravity will act on the center of pole, directly down with a force of . Using basic. trig for a right triangle, the vertical component of that force is , in the y-direction. At the very end of the pole where the string is attached, however, the pole is creating a torque, given by . In this case, (distance from center of pole to the end) and , so the torque from the pole is .

Also in the y-direction is the vertical component of tension, . From right triangle trigonometry, the vertical component of the tension in the rope is . Because the rope is also creating a torque of (length of pole), the force is

From Newton's 2nd Law:

Since we're finding the force required to keep the pole motionless, there should be no acceleration in the y-direction. Therefore, :

Therefore,

your answer is yx

hope this

~mia