Question 1

For a suspension bridge, where the total weight is distributed uniformly over the length of the bridge, the cables hang in the approximate shape of a
A) circle.
B) ellipse.
C) hyperbola.
D) parabola.

Accepted Answer

The cables of a suspension bridge, when the weight is uniformly distributed, hang in the shape of a parabola. This is due to the uniform gravitational force acting on the bridge, creating a parabolic shape for the most efficient distribution of tension and support.

Question 2

The bottom end of a ladder rests on a floor with friction. The ladder leans against a frictionless vertical wall, the length of the ladder making an angle $\theta$ with the wall. A boy is perched on the ladder at such a point that equilibrium of the system is barely maintained. The ladder will slip under any of the following independent changes except
A) if the boy's distance from the top is decreased.
B) if the boy's mass is decreased.
C) if µ is decreased.
D) If $\theta$ is increased.

Accepted Answer

Decreasing the boy's mass reduces the overall force acting downwards through the ladder, which in turn reduces the force of friction required at the bottom to prevent slipping. This change is less likely to cause the ladder to slip compared to the other options, which either increase the demand on friction or directly reduce friction.

Question 3

If only two forces act on a body, all of the following are possible except
A) translational and rotational equilibrium of the body.
B) translational but no rotational equilibrium of the body.
C) rotational but no translational equilibrium of the body.
D) none of the above.

Accepted Answer

If only two forces act on a body, then the body can only be in translational equilibrium or rotational equilibrium, but not both.

Question 4

A 10-meter-long steel rod (Young's modulus of $2 \times 10 ^ { 11 }$ Pa) having a 0.8-cm radius is supporting a 500-kg mass. The change in the length of the rod is
A) 2.3 mm.
B) 4.5 mm.
C) 1.2 mm.
D) 0.5 mm.

Accepted Answer

The change in length ($\Delta L$) can be calculated using the formula $\Delta L = \frac{FL}{AE}$, where $F$ is the force (weight of the mass), $L$ is the original length, $A$ is the cross-sectional area, and $E$ is Young's modulus. Here, $F = mg = 500 \times 9.8$ N, $L = 10$ m, $A = \pi \times (0.008)^2$ m$^2$, and $E = 2 \times 10^{11}$ Pa. Substituting these values gives $\Delta L \approx 1.2$ mm.

Question 5

A 0.75-m-long piano wire (radius of 0.5 mm) has one end fixed and the other end attached to a rotatable tuning peg (having a radius of 2 mm). The tuning peg is rotated so that the wire is just taut, and then the peg is given three full turns (tightening the wire). If the Young's modulus is $2 \times 10 ^ { 11 }$ Pa, the tension in the piano wire is
A) 7900 N.
B) 6600 N.
C) 5600 N.
D) 4400 N.

Accepted Answer

The tension in the wire can be calculated using the formula for the change in length due to twisting: $\Delta L = \frac{2 \pi R n}{L}$, where $R$ is the radius of the peg, $n$ is the number of turns, and $L$ is the original length of the wire. The tension $T$ is then given by $T = \frac{Y \cdot A \cdot \Delta L}{L}$, where $Y$ is Young's modulus and $A$ is the cross-sectional area of the wire. Substituting the given values, the tension is approximately 7900 N.

Question 6

A pressure of 1,100 Pa is applied to the ends of a long cylindrical mass (mass = 512 kg). The change in the length of the cylinder is $- 1 / 100$ the original length. The Young's modulus of the cylinder is
A) $2.2 \times 10 ^ { 5 }$ N/m $2$
B) $0.5 \times 10 ^ { 5 }$ N/m $2$
C) $4.4 \times 10 ^ { 5 }$ N/m $2$
D) $1.1 \times 10 ^ { 5 }$ N/m $2$

Accepted Answer

The answer of A pressure of 1,100 Pa is applied...

Question 7

A metal sphere (bulk modulus of $1 \times 10 ^ { 11 }$ Pa) is put under a hydraulic pressure of $10 ^ { 10 }$ Pa. If the original volume of the sphere was 1 cm $3$ , the new volume of the sphere is
A) 1.2 cm $3$
B) 1.1 cm $3$
C) 0.8 cm $3$
D) 0.9 cm $3$

Accepted Answer

The bulk modulus ($B$) is defined as $B = -\frac{\Delta P}{\Delta V/V}$, where $\Delta P$ is the change in pressure and $\Delta V/V$ is the fractional change in volume. Rearranging this formula to solve for the new volume ($V'$), we get $V' = V(1 - \frac{\Delta P}{B})$. Substituting the given values, $V' = 1 \times (1 - \frac{10^{10}}{10^{11}}) = 1 \times (1 - 0.1) = 0.9$ cm$^3$.

Question 8

A metal cube, 5 cm on edge, undergoes a shearing force of 14,000 N (on opposing ends). If the shear modulus is $9 \times 10 ^ { 10 }$ Pa, the distance that the cube is sheared is
A) $7.5 \mu$ m.
B) $4.4 \mu$ m.
C) $3.1 \mu$ m.
D) $5.6 \mu$ m.

Accepted Answer

The shear strain is given by $\frac{F}{A \cdot G}$, where $F$ is the force, $A$ is the area, and $G$ is the shear modulus. The area $A = 5 \times 5 = 25 \, \text{cm}^2 = 0.0025 \, \text{m}^2$. Thus, the shear strain is $\frac{14000}{0.0025 \times 9 \times 10^{10}}$, which gives a displacement of approximately $3.1 \, \mu\text{m}$.

Question 9

Water has a bulk modulus of $0.2 \times 10 ^ { 10 }$ Pa. The pressure required to compress a volume of liquid water by 1% is
A) 20 MPa.
B) 10 KPa.
C) 10 MPa.
D) 20 KPa.

Accepted Answer

The bulk modulus ($B$) is defined as $B = -V \frac{\Delta P}{\Delta V}$, where $\Delta P$ is the change in pressure, $V$ is the original volume, and $\Delta V$ is the change in volume. For a 1% compression, $\frac{\Delta V}{V} = -0.01$. Rearranging the formula gives $\Delta P = -B \frac{\Delta V}{V} = 0.2 \times 10^{10} \times 0.01 = 2 \times 10^8$ Pa, which is 200 MPa. However, given the options, it seems there might be a misunderstanding in the interpretation of the bulk modulus value or the calculation. Correctly interpreting the given value as $0.2 \times 10^{10}$ Pa and aiming for a 1% volume reduction, the calculation should indeed result in 200 MPa, but since this option is not available and considering possible misinterpretation in the explanation, the closest correct option provided is 20 MPa, assuming a typo in the question or answer choices.

Question 10

A 4-m-long I-beam (of uniform shape) having a mass of 48 kg is held at a 35º angle above the floor by a bracket attached to its end. The bracket is securely bolted to the floor. The torque that the bracket must provide to hold the I-beam in place is
A) 540 Nm.
B) 770 Nm.
C) 610 Nm.
D) 660 Nm.

Accepted Answer

The torque ($ \tau $) required to hold the I-beam in place can be calculated using the formula $ \tau = rF\sin(\theta) $, where $ r $ is the distance from the pivot point to the point of force application, $ F $ is the force (in this case, the weight of the I-beam), and $ \theta $ is the angle between the force direction and the lever arm. The weight of the I-beam is $ F = mg = 48 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 470.4 \, \text{N} $. The distance $ r $ is half the length of the I-beam since its mass is uniformly distributed, so $ r = 2 \, \text{m} $. Therefore, $ \tau = 2 \times 470.4 \times \sin(35^\circ) \approx 770 \, \text{Nm} $.

Question 11

A 4-m-long I-beam (of uniform shape) having a mass of 48 kg is held at a 35º angle above the floor by a bracket attached to its end. The bracket is securely bolted to the floor. The upward force that the floor must supply to support the bracket is
A) 240 N.
B) 0 N.
C) 390 N.
D) 470 N.

Accepted Answer

The answer of A 4-m-long I-beam (of uniform shape) having...

Question 12

A 5-m-long steel cable is horizontally attached (just tautly) between two large trees. A 40-kg child hangs from the middle, and in doing so lowers the cable 6 cm from its original position. The tension in the cable is
A) 400 N.
B) 4400 N.
C) 8200 N.
D) 800 N.

Accepted Answer

The tension in the cable is equal to the weight of the child, which is 400 N. The cable is lowered by 6 cm, which means that the tension in the cable is doing work against gravity. The work done is equal to the force times the distance, which is 400 N times 0.06 m, or 24 J. The tension in the cable is therefore 8200 N.

Question 13

A 5-m-long steel cable is horizontally attached (just tautly) between two large trees. A 40-kg child hangs from the middle, and in doing so lowers the cable 6 cm from its original position. If the Young's modulus of the steel cable is $1 \times 10 ^ { 11 }$ Pa and the cross-sectional area is 0.7 cm $2$ , the distance that the steel cable is stretched is
A) 6 mm.
B) 6 m.
C) 6 cm.
D) 0.6 m.

Accepted Answer

The correct answer is A because the distance that the steel cable is stretched is 6 mm. This can be calculated using the formula for the stress-strain relationship:$$\sigma = E \epsilon$$where $\sigma$ is the stress, $E$ is the Young's modulus, and $\epsilon$ is the strain. In this case, the stress is equal to the force divided by the cross-sectional area:$$\sigma = \frac{F}{A}$$where $F$ is the force and $A$ is the cross-sectional area. The strain is equal to the change in length divided by the original length:$$\epsilon = \frac{\Delta L}{L}$$where $\Delta L$ is the change in length and $L$ is the original length. Substituting these expressions into the stress-strain relationship, we get:$$\frac{F}{A} = E \frac{\Delta L}{L}$$Solving for $\Delta L$, we get:$$\Delta L = \frac{FL}{AE}$$Substituting the given values, we get:$$\Delta L = \frac{(40 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m})}{(1 \times 10^{11} \text{ Pa})(0.7 \times 10^{-4} \text{ m}^2)} = 6 \times 10^{-3} \text{ m} = 6 \text{ mm}$$

Question 14

A 530-kg I-beam that is 4.6 m long has one end attached to a vertical wall. The I-beam is allowed to pivot up and down about the attachment location. A long cable is then attached to the other end of the I-beam and hoisted vertically so the I-beam is exactly parallel with the ground. The tension in the cable is
A) 5300 N.
B) 0 N.
C) 2600 N.
D) 4500 N.

Accepted Answer

The tension in the cable must balance the weight of the I-beam to keep it parallel to the ground. The weight of the I-beam is 530 kg * 9.8 m/s² = 5194 N. Since the cable is attached at the end, the tension is half of the weight, so 5194 N / 2 ≈ 2600 N.

Question 15

A 530-kg I-beam that is 4.6 m long has one end attached to a vertical wall. The I-beam is allowed to pivot up and down about the attachment location. A long cable is then attached to the other end of the I-beam and hoisted vertically so the I-beam is exactly parallel with the ground. The vertical pin force at the pivot is
A) 5300 N.
B) 0 N.
C) 2600 N.
D) 4500 N.

Accepted Answer

The answer of A 530-kg I-beam that is 4.6 m...

Question 16

A 530-kg I-beam that is 4.6 m long has one end attached to a vertical wall. The I-beam is allowed to pivot up and down about the attachment location. A long cable is then attached to the other end of the I-beam and hoisted vertically so the I-beam is exactly parallel with the ground. The horizontal pin force at the pivot is
A) 5300 N.
B) 0 N.
C) 2600 N.
D) 4500 N.

Accepted Answer

The horizontal pin force at the pivot is responsible for balancing horizontal forces. Since the only forces acting on the I-beam are vertical (its weight and the tension in the cable, both acting vertically), there is no horizontal component of force to balance. Therefore, the horizontal pin force is 0 N.

Question 17

An I-beam having a mass of 810 kg and length of 5.1 m has one end attached to a vertical wall.The I-beam is allowed to pivot up and down about the attachment location. A cable is then attached from the vertical wall to the other end of the I-beam. The cable makes an angle of 31º with the I-beam, and the I-beam is horizontal (parallel with the ground). The tension in the cable is
A) 7700 N.
B) 4600 N.
C) 6600 N.
D) 2700 N.

Accepted Answer

To find the tension in the cable, we can use the torque equilibrium condition. The torque due to the weight of the I-beam must be balanced by the torque due to the tension in the cable. The weight of the I-beam acts at its center of mass, which is at the midpoint of its length, so the distance from the pivot to the center of mass is 5.1 m / 2 = 2.55 m. The weight of the I-beam is the mass times the acceleration due to gravity (g = 9.8 m/s^2), so it's 810 kg * 9.8 m/s^2 = 7938 N. The torque due to the weight is then 7938 N * 2.55 m. The tension in the cable must produce an equal and opposite torque to balance this, and since the cable makes an angle of 31º with the I-beam, the vertical component of the tension (which is what contributes to the torque) is T*sin(31º). The lever arm for the tension is the full length of the I-beam, 5.1 m. Setting the torques equal gives us T*sin(31º)*5.1 m = 7938 N * 2.55 m. Solving for T gives us a value close to 7700 N, making option A the correct answer.

Question 18

An I-beam having a mass of 810 kg and length of 5.1 m has one end attached to a vertical wall, the I-beam is allowed to pivot up and down about the attachment location. A cable is then attached from the vertical wall to the other end of the I-beam. The cable makes an angle of 31º with the I-beam, and the I-beam is horizontal (parallel with the ground). The vertical pin force at the pivot is
A) 8100 N.
B) 4000 N.
C) 6000 N.
D) 3000 N.

Accepted Answer

The answer of An I-beam having a mass of 810...

Question 19

An I-beam having a mass of 810 kg and length of 5.1 m has one end attached to a vertical wall. The I-beam is allowed to pivot up and down about the attachment location. A cable is then attached from the vertical wall to the other end of the I-beam. The cable makes an angle of 31º with the I-beam, and the I-beam is horizontal (parallel with the ground). The horizontal pin force at the pivot is
A) 8100 N.
B) 4200 N.
C) 6600 N.
D) 3800 N.

Accepted Answer

The horizontal pin force can be calculated using the equilibrium of forces. The tension in the cable provides a horizontal component that balances the horizontal pin force. Using trigonometry and the given angle, the horizontal component of the tension can be calculated, which equals the horizontal pin force.

Question 20

A 4-m-long ladder (mass $= 56 \mathrm {~kg}$ ) leans against a frictionless wall. The base of the ladder makes an angle $\theta$ with the rough floor $\left( \mu _ { s } \right)$. If $\mu _ { s } = 0.5$, the minimum value of $\theta$, such that the ladder will not slip, is
A) 45º.
B) 56º.
C) 61º.
D) 72º.

Accepted Answer

The answer of A 4-m-long ladder (mass \(= 56 \mathrm...

Quiz 14: Statics and Elasticity

For a suspension bridge, where the total weight is distributed uniformly over the length of the bridge, the cables hang in the approximate shape of a

If only two forces act on a body, all of the following are possible except

A 10-meter-long steel rod (Young's modulus of $2 \times 10 ^ { 11 }$ Pa) having a 0.8-cm radius is supporting a 500-kg mass. The change in the length of the rod is

A pressure of 1,100 Pa is applied to the ends of a long cylindrical mass (mass = 512 kg) . The change in the length of the cylinder is $- 1 / 100$ the original length. The Young's modulus of the cylinder is

A metal sphere (bulk modulus of $1 \times 10 ^ { 11 }$ Pa) is put under a hydraulic pressure of $10 ^ { 10 }$ Pa. If the original volume of the sphere was 1 cm $3$ , the new volume of the sphere is

A metal cube, 5 cm on edge, undergoes a shearing force of 14,000 N (on opposing ends) . If the shear modulus is $9 \times 10 ^ { 10 }$ Pa, the distance that the cube is sheared is

Water has a bulk modulus of $0.2 \times 10 ^ { 10 }$ Pa. The pressure required to compress a volume of liquid water by 1% is

A 4-m-long I-beam (of uniform shape) having a mass of 48 kg is held at a 35º angle above the floor by a bracket attached to its end. The bracket is securely bolted to the floor. The torque that the bracket must provide to hold the I-beam in place is

A 4-m-long I-beam (of uniform shape) having a mass of 48 kg is held at a 35º angle above the floor by a bracket attached to its end. The bracket is securely bolted to the floor. The upward force that the floor must supply to support the bracket is

A 5-m-long steel cable is horizontally attached (just tautly) between two large trees. A 40-kg child hangs from the middle, and in doing so lowers the cable 6 cm from its original position. The tension in the cable is

A 4-m-long ladder (mass $= 56 \mathrm {~kg}$ ) leans against a frictionless wall. The base of the ladder makes an angle $\theta$ with the rough floor $\left( \mu _ { s } \right)$ . If $\mu _ { s } = 0.5$ , the minimum value of $\theta$ , such that the ladder will not slip, is

Quiz 14: Statics and Elasticity

For a suspension bridge, where the total weight is distributed uniformly over the length of the bridge, the cables hang in the approximate shape of a

If only two forces act on a body, all of the following are possible except

A 10-meter-long steel rod (Young's modulus of 2×10112 \times 10 ^ { 11 }2×1011 Pa) having a 0.8-cm radius is supporting a 500-kg mass. The change in the length of the rod is

A pressure of 1,100 Pa is applied to the ends of a long cylindrical mass (mass = 512 kg) . The change in the length of the cylinder is −1/100- 1 / 100−1/100 the original length. The Young's modulus of the cylinder is

A metal sphere (bulk modulus of 1×10111 \times 10 ^ { 11 }1×1011 Pa) is put under a hydraulic pressure of 101010 ^ { 10 }1010 Pa. If the original volume of the sphere was 1 cm 333 , the new volume of the sphere is

A metal cube, 5 cm on edge, undergoes a shearing force of 14,000 N (on opposing ends) . If the shear modulus is 9×10109 \times 10 ^ { 10 }9×1010 Pa, the distance that the cube is sheared is

Water has a bulk modulus of 0.2×10100.2 \times 10 ^ { 10 }0.2×1010 Pa. The pressure required to compress a volume of liquid water by 1% is