Question

1. une grue maintient en équilibre un bloc de métal de 10 000 kg. étant donné que la masse du bras de la grue est de 1000 kg et que sa longueur est de 10,0 m, quelle est la tension dans le câble a et quelle est la grandeur de la force exercée par le pivot sur le bras de la grue ? (faites un diagramme des forces séparé du dessin.) 15 points (image of a crane with angles 10.0° and 45.0°, cables a and b, and a pivot)

Answer:

.0^\circ = 55.0^\circ \) above the negative x-axis? Wait, this is getting confusing. Maybe better to use the torque result first.

Alternatively, since we have \( T_a \approx 4.20 \times 10^5 \, \text{N} \), \( T_b = m_{block}g = 98100 \, \text{N} \), weight of arm \( F_{arm} = m_{arm}g = 9810 \, \text{N} \).

Forces in x-direction: \( F_{p,x} = T_a \cos(10.0^\circ) - T_b \cos(45.0^\circ) \) (since \( T_a \) has a horizontal component to the right, \( T_b \) has a horizontal component to the left (because the block is hanging, so the rope \( b \) is vertical? Wait, no, rope \( b \) is vertical, so it has no horizontal component. Oh! Wait, the block is hanging vertically, so rope \( b \) is vertical, so \( T_b \) is vertical (downward? No, upward? Wait, the block is in equilibrium, so \( T_b \) is upward, balancing the block's weight. So \( T_b = m_{block}g \) upward. Then the arm is at an angle, so the end of the arm has a vertical rope \( b \) with tension \( T_b = m_{block}g \) upward. Wait, no, the block is hanging, so \( T_b \) is upward (tension in rope \( b \) is upward, balancing the block's weight downward). So the forces on the arm:

• Tension \( T_a \) from cable \( a \), at angle \( 10.0^\circ \) to the arm (so the angle between \( T_a \) and the arm is \( 10.0^\circ \), so if the arm is at angle \( \theta = 45.0^\circ \) from vertical (i.e., \( 45.0^\circ \) from y-axis), then \( T_a \) is at \( \theta + 10.0^\circ \) from y-axis? No, the diagram shows cable \( a \) is attached to the crane body and the arm, making \( 10.0^\circ \) with the arm. So the arm is a lever, pivot at the crane body. So the arm is a rigid rod, pivot at the crane (the pivot point). The cable \( a \) is attached to the end of the arm and the crane body, making \( 10.0^\circ \) with the arm. The rope \( b \) is attached to the end of the arm and the block, hanging vertically. So the forces on the arm are:

1. Tension \( T_a \) from cable \( a \), acting at the end of the arm, at an angle of \( 10.0^\circ \) above the arm (so the angle between \( T_a \) and the arm is \( 10.0^\circ \), so the direction of \( T_a \) is \( 10.0^\circ \) from the arm, towards the crane body).

1. Tension \( T_b \) from rope \( b \), acting at the end of the arm, vertically downward (since the block is pulling down on the arm).

1. Weight of the arm \( F_{arm} = m_{arm}g \), acting at the center of the arm, vertically downward.

1. Force from the pivot \( \vec{F}_p \), acting at the pivot point (crane body), with components \( F_{p,x} \) (horizontal) and \( F_{p,y} \) (vertical).

Now, for torque about the pivot: the torque due to \( T_a \) is counterclockwise (positive), torque due to \( T_b \) and \( F_{arm} \) is clockwise (negative).

The length of the arm is \( L = 10.0 \, \text{m} \). The perpendicular distance from pivot to \( T_a \) is \( L \sin(10.0^\circ) \) (since the angle between \( T_a \) and the arm is \( 10.0^\circ \), so the perpendicular component of \( T_a \) is \( T_a \sin(10.0^\circ) \), and torque is \( r \times F = L \times T_a \sin(10.0^\circ) \)).

The perpendicular distance from pivot to \( T_b \) is \( L \cos(45.0^\circ) \) (since the arm is at \( 45.0^\circ \) from vertical, so the horizontal distance from pivot to the end of the arm is \( L \sin(45.0^\circ) \)? Wait, no, torque is \( r \times F \), where \( r \) is the position vector from pivot to the point of application, and \( F \) is the force. For \( T_b \), which is vertical (downward), the position vector is \( L \) at an angle of \( 45.0^\circ \) from vertical (i.e., \( 45.0^\circ \) from y-axis), so the angle between \( r \) and \( F \) (which is downward, along -y) is \( 45.0^\circ \), so the torque is \( r F \sin(45.0^\circ) = L T_b \sin(45.0^\circ) \) (clockwise, so negative).

For \( F_{arm} \), the position vector is \( L/2 \) at \( 45.0^\circ \) from y-axis, so torque is \( (L/2) F_{arm} \sin(45.0^\circ) \) (clockwise, negative).

So torque equilibrium:

\( T_a L \sin(10.0^\circ) - T_b L \sin(45.0^\circ) - F_{arm} (L/2) \sin(45.0^\circ) = 0 \)

Divide both sides by \( L \):

\( T_a \sin(10.0^\circ) - T_b \sin(45.0^\circ) - (F_{arm}/2) \sin(45.0^\circ) = 0 \)

We know \( T_b = m_{block}g = 10000 \times 9.81 = 98100 \, \text{N} \), \( F_{arm} = m_{arm}g = 1000 \times 9.81 = 9810 \, \text{N} \)

So:

\( T_a \sin(10.0^\circ) = T_b \sin(45.0^\circ) + (F_{arm}/2) \sin(45.0^\circ) \)

\( T_a = \frac{ \sin(45.0^\circ) (T_b + F_{arm}/2) }{ \sin(10.0^\circ) } \)

Plug in the numbers:

\( \sin(45.0^\circ) \approx 0.7071 \), \( \sin(10.0^\circ) \approx 0.1736 \), \( T_b = 98100 \, \text{N} \), \( F_{arm}/2 = 4905 \, \text{N} \)

Numerator: \( 0.7071 \times (98100 + 4905) = 0.7071 \times 103005 \approx 72825 \)

Denominator: \( 0.1736 \)

\( T_a \approx 72825 / 0.1736 \approx 419500 \, \text{N} \approx 4.20 \times 10^5 \, \text{N} \) (same as before)

Now, for the pivot force:

Forces in x-direction: \( F_{p,x} = T_a \cos(10.0^\circ) \) (since \( T_b \) is vertical, no x-component; the arm is in equilibrium, so horizontal forces must balance. \( T_a \) has a horizontal component to the right, so \( F_{p,x} \) must be to the left? Wait, no: the pivot exerts a force on the arm, so if \( T_a \) is pulling the arm to the right (horizontal component), then the pivot must pull the arm to the left to balance. Wait, no, the arm is a rigid body, so the forces on the arm are: \( T_a \) (from cable \( a \), pulling the arm towards the crane body, so direction of \( T_a \) is from the end of the arm to the crane body, so angle \( 10.0^\circ \) above the arm. The arm is at \(