Question

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

Answer 1

Answer:

$397.44\text{ N}}$ or approximately $4.0\cdot 10^2\text{ N}$ 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 $F_g$. Using basic. trig for a right triangle, the vertical component of that force is $F_g\cos 30^{\circ}$, 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 $\tau =rF\cos\theta$. In this case, $r=3.75$ (distance from center of pole to the end) and $\theta = 30^{\circ}$, so the torque from the pole is $3.75\cdot F_g\cos 30^{\circ}$.

Also in the y-direction is the vertical component of tension, $T$. From right triangle trigonometry, the vertical component of the tension in the rope is $T\sin 20^{\circ}$. Because the rope is also creating a torque of $r=7.5$ (length of pole), the force is $7.5\cdot T\sin 20^{\circ}$

From Newton's 2nd Law:

$\displaystyle \sum F_y=ma_y$

Since we're finding the force required to keep the pole motionless, there should be no acceleration in the y-direction. Therefore, $\displaystyle \sum F_y=ma_y=0$:

$\displaystyle \sum F_y=7.5T\sin20^{\circ}-3.75F_g\cos 30^{\circ}=0,\\\\7.5T\sin20^{\circ}=3.75F_g\cos 20^{\circ},\\\\T=\frac{3.75F_g\cos20^{\circ}}{7.5\sin 20^{\circ}}$

$F_g=m_{pole}g=32\cdot 9.81=313.92$

Therefore,

$\displaystyle T=\frac{313.92\cos30^{\circ}}{2\sin20^{\circ}}=\boxed{397.44\text{ N}}}$