Question

1. two masses (m1 = 3.00 kg, m2 = 5.00 kg) hang from the ends of a metre - stick as shown. if the mass of the metre - stick is negligible, at what distance from the left of the metre - stick should a pivot be placed so that the system will be balanced?
2. a uniform 400 n diving board is supported at two points as shown in the diagram. if a 75 kg diver stands at the end of the board, what are the forces acting on the each support?
3. a 650 n student stands on a 250 n uniform beam that is supported by two supports as shown in the diagram. if the supports are 5.0 m apart and the student stands 1.5 m from the left support:

a) what is the force that the right support exerts on the beam?
b) what is the force that the left support exerts on the beam?

1. what is the upward force, f, in the diagram below?

answers: 1. f = 5.3n 2. d = 0.37m 4. t = 200n 5. m = 93kg 6. t = 10.5n 7. d = 0.625m 8. left = 265n ↓ right 3740n ↑ 9a) right - 320n↑ b) left 580n↑ 10. f = 608n.

Explanation:

Step 1: Set up torque - balance equation for problem 7

Let the distance of the pivot from the left - hand side be \(x\) (in meters). The torque \(\tau = rF\). For the system to be in balance, the torques on either side of the pivot must be equal. The force due to \(m_1\) is \(F_1=m_1g\) and the force due to \(m_2\) is \(F_2 = m_2g\). The distance of \(m_1\) from the pivot is \(x\) and the distance of \(m_2\) from the pivot is \(1 - x\). So, \(m_1g\times x=m_2g\times(1 - x)\). Since \(g\) cancels out, we have \(m_1x=m_2(1 - x)\).

Step 2: Solve for \(x\)

Expand the right - hand side: \(m_1x=m_2 - m_2x\). Rearrange the terms: \(m_1x+m_2x=m_2\). Factor out \(x\): \(x(m_1 + m_2)=m_2\). Then \(x=\frac{m_2}{m_1 + m_2}\). Substitute \(m_1 = 3.00\ kg\) and \(m_2 = 5.00\ kg\) into the formula: \(x=\frac{5}{3 + 5}=\frac{5}{8}=0.625\ m\).

Step 3: Set up equations for problem 8

Let the left - hand support force be \(F_1\) and the right - hand support force be \(F_2\). The weight of the diving board \(W_{board}=400\ N\) and the weight of the diver \(W_{diver}=mg=75\times9.8 = 735\ N\). Take the torque about the left - hand support. The weight of the board acts at its center of mass. The length of the board is \(L = 2.0\ m\) and the distance between the supports is \(d = 0.5\ m\). The equation for torque balance \(\sum\tau=0\) gives \(W_{diver}\times2+W_{board}\times1=F_2\times0.5\). Also, from the force balance in the vertical direction \(F_1+F_2=W_{diver}+W_{board}\).
First, from the torque equation: \(735\times2 + 400\times1=F_2\times0.5\). \(1470+400 = 0.5F_2\), \(1870 = 0.5F_2\), so \(F_2=\frac{1870}{0.5}=3740\ N\). Then from the force balance \(F_1=(735 + 400)-3740=-2605\ N\) (the negative sign just indicates the direction, and the magnitude is \(2605\ N\)).

Step 4: Set up equations for problem 9

Let the force of the left - hand support be \(F_1\) and the force of the right - hand support be \(F_2\). The weight of the student \(W_{student}=650\ N\) and the weight of the beam \(W_{beam}=250\ N\). The length between the supports is \(L = 5.0\ m\) and the student is at a distance \(x = 1.5\ m\) from the left - hand support.
Take the torque about the left - hand support: \(W_{student}\times1.5+W_{beam}\times2.5=F_2\times5\). \(650\times1.5+250\times2.5=F_2\times5\). \(975 + 625=5F_2\), \(1600 = 5F_2\), so \(F_2 = 320\ N\). From the force balance \(F_1+F_2=W_{student}+W_{beam}\), so \(F_1=(650 + 250)-320=580\ N\).

Step 5: Set up equation for problem 10

Take the torque about the left - hand end. Let the upward force be \(F\). The weight of the first part is \(W_1 = 270\ N\) at a distance \(d_1 = 12\ m\), the weight of the second part is \(W_2 = 1080\ N\) at a distance \(d_2 = 24\ m\), and the force \(F\) is at a distance \(d_3 = 48\ m\). The torque balance equation \(\sum\tau=0\) gives \(270\times12+1080\times24=F\times48\). \(3240+25920 = 48F\), \(29160 = 48F\), so \(F = 608\ N\).

Answer:

1. \(0.625\ m\)
2. Left: \(2605\ N\), Right: \(3740\ N\)
3. a) \(320\ N\), b) \(580\ N\)
4. \(608\ N\)