Question

worksheet #2 - torque @ 90°

1. what force f is needed to balance the beam in the diagram below?

0 cm 50 cm 100 cm 175 cm
8.0 n pivot cg f

1. how far from the pivot must the 64 n object be placed to balance the beam in the diagram below?

42 cm cg x =?
56 n pivot 64 n

1. if the torque needed to loosen a lug nut is 45 nm and you are using a 35 cm wheel wrench, what force do you need to exert perpendicular to the end of the wrench?
2. a beam of negligible mass is attached to a wall by a hinge. attached to the center of the beam is a 400 n weight. a rope supports the beam as shown in the diagram. what is the tension in the rope?

t
400 n

1. two students sit on either end of a uniform teeter - totter. student 1 sits 1.10 m from the pivot while student 2 sits 0.85 m from the pivot. if student 1 has a mass of 72 kg, what is the mass of student 2?

1.2 m

1. a 0.75 kg bird stands on a uniform 1.0 kg stick as shown. the stick is attached to a wall with a hinge and to the ceiling with a rope of negligible mass. what is the tension in the rope?

0.90 m

Explanation:

Step1: Recall torque - balance formula

For a balanced beam, the sum of the torques about the pivot point is zero, $\sum\tau = 0$. Torque $\tau= rF\sin\theta$, and when $\theta = 90^{\circ}$, $\sin\theta = 1$, so $\tau=rF$.

Step2: Solve for the first - question

The torque due to the $8.0\ N$ force about the pivot is $\tau_1 = 8.0\ N\times(100 - 50)\ cm=8.0\ N\times0.5\ m$. Let the force $F$ act at a distance $r_2=(175 - 100)\ cm = 0.75\ m$ from the pivot. Setting $\tau_1=\tau_2$, we have $8.0\ N\times0.5\ m=F\times0.75\ m$. Then $F=\frac{8.0\ N\times0.5\ m}{0.75\ m}=\frac{4}{0.75}\ N\approx5.33\ N$.

Step3: Solve for the second - question

Let the distance of the $64\ N$ object from the pivot be $x$. The torque due to the $56\ N$ force about the pivot is $\tau_1 = 56\ N\times42\ cm$. The torque due to the $64\ N$ force about the pivot is $\tau_2=64\ N\times x$. Setting $\tau_1 = \tau_2$, we get $56\ N\times42\ cm=64\ N\times x$. Then $x=\frac{56\ N\times42\ cm}{64\ N}=36.75\ cm$.

Step4: Solve for the third - question

Given $\tau = 45\ Nm$ and $r = 35\ cm=0.35\ m$. Since $\tau=rF$ (because $\theta = 90^{\circ}$), then $F=\frac{\tau}{r}=\frac{45\ Nm}{0.35\ m}\approx128.57\ N$.

Step5: Solve for the fourth - question

Let the length of the beam be $L$. Taking the torque about the hinge, the weight of $400\ N$ acts at the center of the beam. Let the tension in the rope be $T$. The distance from the hinge to the point where the tension acts is $L$ and the distance from the hinge to the center of the beam is $\frac{L}{2}$. Setting the sum of torques about the hinge equal to zero, $\sum\tau=0$. The torque due to the weight is $\tau_w = 400\ N\times\frac{L}{2}$ and the torque due to the tension is $\tau_T=T\times L$. Then $400\ N\times\frac{L}{2}=T\times L$, and $T = 200\ N$.

Step6: Solve for the fifth - question

For the teeter - totter, taking torques about the pivot. The weight of Student 1 is $F_1=m_1g$ and the weight of Student 2 is $F_2=m_2g$. The distance of Student 1 from the pivot is $r_1 = 1.10\ m$ and the distance of Student 2 from the pivot is $r_2=0.85\ m$. Since $\sum\tau = 0$, $m_1g\times r_1=m_2g\times r_2$. Canceling out $g$, we have $m_2=\frac{m_1r_1}{r_2}=\frac{72\ kg\times1.10\ m}{0.85\ m}\approx93.88\ kg$.

Step7: Solve for the sixth - question

The weight of the stick is $W_s=m_sg$ where $m_s = 1.0\ kg$ and $g = 9.8\ m/s^2$, and the weight of the bird is $W_b=m_bg$ where $m_b = 0.75\ kg$. Taking torques about the hinge. Let the length of the stick be $L = 1.2\ m$ and the distance of the bird from the hinge be $r_b=0.90\ m$, and the center of mass of the stick is at $r_s = 0.6\ m$. The tension in the rope is $T$. The sum of torques about the hinge $\sum\tau=0$. The torque due to the weight of the stick is $W_s\times0.6\ m$, the torque due to the weight of the bird is $W_b\times0.9\ m$, and the torque due to the tension is $T\times1.2\ m$. So $m_sg\times0.6\ m+m_bg\times0.9\ m=T\times1.2\ m$. Substituting $m_s = 1.0\ kg$, $m_b = 0.75\ kg$ and $g = 9.8\ m/s^2$:
\[

$$\begin{align*} (1.0\ kg\times9.8\ m/s^2\times0.6\ m)+(0.75\ kg\times9.8\ m/s^2\times0.9\ m)&=T\times1.2\ m\\ (9.8\times0.6)\ Nm+(9.8\times0.75\times0.9)\ Nm&=T\times1.2\ m\\ 5.88\ Nm + 6.615\ Nm&=T\times1.2\ m\\ 12.495\ Nm&=T\times1.2\ m\\ T&=\frac{12.495\ Nm}{1.2\ m}\approx10.41\ N \end{align*}$$

\]

Answer:

1. $F\approx5.33\ N$
2. $x = 36.75\ cm$
3. $F\approx128.57\ N$
4. $T = 200\ N$
5. $m_2\approx93.88\ kg$
6. $T\approx10.41\ N$