Physics- Ch. 8 Test Review

A 73 kg mass is suspended from two identical cables, each at an angle of 62° from the vertical. Which statement correctly describes the magnitudes of the forces F₁ and F₂ exerted by the respective cables upon the hanging mass?

• F₁ and F₂ are equally strong, and each has a magnitude that is less than 358 N.

• F₁ is greater than F₂.

• F₁ and F₂ are equally strong, and each has a magnitude that is greater than 358 N.

• F₂ is greater than F₁.

• These cables will be unable to support the mass.

Answer explanation

A 73 kg mass is suspended from two identical cables, each at an angle of 62° from the vertical. Which statement correctly describes the magnitudes of the forces F₁ and F₂ exerted by the respective cables upon the hanging mass?

Resolving the tensions in the cables into components, we see that F₁ and F₂ must be equal by symmetry, and that both must be greater than half of the weight of the mass (358 N), since only the vertical component in each cable contributes to balancing out the weight.

Tags

NGSS.HS-PS2-1

A cable is used to lift a 8.00×10⁵ kg container from a ship. The tensile strength of the cable is 4.30×10¹⁰ N/m². With a safety factor of 3, what is the minimum cross sectional area required for a cable that can safely support the container?

• 1.82 cm²

• 5.47×10⁴ cm²

• 0.558 cm²

• 0.0547 cm²

• 5.47 cm²

Answer explanation

A cable is used to lift a 8.00×10⁵ kg container from a ship. The tensile strength of the cable is 4.30×10¹⁰ N/m². With a safety factor of 3, what is the minimum cross sectional area required for a cable that can safely support the container?

The tensile force in the cable is related to the tensile strength, σ, and the cable area, A by F=σA, F=σA
With the safety factor equal to 3 the equation becomes:
3F=σA⇒A=3Fσ=3mgσ=38×105kg  (9.8m/s2)4.3×1010N/m2=3F=σA⇒A=3Fσ=3mgσ=38×105kg  (9.8m/s2)4.3×1010N/m2= 5.47×10^-4 m² = 5.47 cm²

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NGSS.HS-PS2-1

NGSS.HS-PS2-4

A 32 kg mass is suspended from two identical cables, each at an angle of 49° from the vertical. Which statement correctly describes the magnitudes of the forces F₁ and F₂ exerted by the respective cables upon the hanging mass?

• F₂ is greater than F₁.

• F₁ and F₂ are equally strong, and each has a magnitude that is greater than 157 N.

• More information is needed.

• F₁ and F₂ are equally strong, and each has a magnitude that is less than 157 N.

• F₁ and F₂ are equally strong, and each has a magnitude that equals 157 N.

Answer explanation

A 32 kg mass is suspended from two identical cables, each at an angle of 49° from the vertical. Which statement correctly describes the magnitudes of the forces F₁ and F₂ exerted by the respective cables upon the hanging mass?

Resolving the tensions in the cables into components, we see that F₁ and F₂ must be equal by symmetry, and that both must be greater than half of the weight of the mass (157 N), since only the vertical component in each cable contributes to balancing out the weight.

Tags

NGSS.HS-PS2-1

A 50 kg painter stands on a scaffold consisting of a 8.0 m long beam with a mass of 20 kg that is supported by posts at each end. What support force must be provided by the left post if the painter stands 2.0 m from the right end?

• 23 N

• 320 N

• 1800 N

• 470 N

• 220 N

Answer explanation

A 50 kg painter stands on a scaffold consisting of a 8.0 m long beam with a mass of 20 kg that is supported by posts at each end. What support force must be provided by the left post if the painter stands 2.0 m from the right end?

Both the sum of forces and torques must be equal in this case. Therefore we have the equations below, where the subscript p refers to the painter and b refers to the beam.

Σ Fy=0=FR+FL−mpg−mbgΣ τ=0=lbFL−12/lbmbg−lpmpg

Σ Fy=0=FR+FL−mpg−mbgΣ τ=0=lbFL−12lbmbg−lpmpg

We can solve for the support force of the left beam using the torque equation.

0=lbFL−12/lbmbg−lpmpgFL=12/lbmbg+lpmpglb=12/(8 m)(20 kg)(9.8 ms2/)+(2 m)(50 kg)(9.8 ms2/)8 

Tags

NGSS.HS-PS2-1

A torque of 7.8 N m is applied by twisting a screwdriver handle. How much force is applied tangentially to the screwdriver if the diameter of the handle is 3.3 cm?

• 240 N

• 26 N

• 4.7 N

• 0.42 N

• 470 N

Answer explanation

A torque of 7.8 N m is applied by twisting a screwdriver handle. How much force is applied tangentially to the screwdriver if the diameter of the handle is 3.3 cm?

The diameter of the screwdriver’s handle is 3.3 cm, therefore the radius is 1.65 cm.
The torque is given by τ = rF, therefore:

F=τ/r= (7.80 N m) / (0.0165 m) = 473 N m

Which of the following statements correctly describes the forces and torques acting on an object in static equilibrium?

• Both the net force and the net torque must be zero.

• The net torque must be less than the net force.

• The net force on the object must be zero, but the net torque must be nonzero.

• Neither the net force nor the net torque can be zero.

• The net force be equal and opposite in value to the net torque.

Answer explanation

Which of the following statements correctly describes the forces and torques acting on an object in static equilibrium?

A static object is at rest with zero acceleration. The net force is zero in all directions, including rotational forces around any center. Therefore, net torque is also zero.

A lightweight beam, 9.0 m long, is supported by a fulcrum at its midpoint. A 4.0 kg mass rests on the beam's left end, and a 7.0 kg mass rests on the right end. A third object, of mass m, is placed a distance d to the left of the fulcrum in order to level the beam. Which mass-and-distance pair will do this?

• m = 8.3 kg and d = 1.23 m

• m = 2.2 kg and d = 0.614 m

• m = 11 kg and d = 1.23 m

• m = 8.8 kg and d = 0.245 m

• m = 11 kg and d = 0.614 m

Answer explanation

A lightweight beam, 9.0 m long, is supported by a fulcrum at its midpoint. A 4.0 kg mass rests on the beam's left end, and a 7.0 kg mass rests on the right end. A third object, of mass m, is placed a distance d to the left of the fulcrum in order to level the beam. Which mass-and-distance pair will do this?

The beam is in static equilibrium, so the sum of the torques about the fulcrum must be equal to zero. This gives

∑τ=0=(A2 m)(4 kg)(g)−(A2 m)(7 kg)(g)−(d)(m)(g)∑τ=0=(A2 m)(4 kg)(g)−(A2 m)(7 kg)(g)−(d)(m)(g), and solving for the product dm, we get dm=(9 m)(7−4 kg)2=dm=(9 m)(7−4 kg)2=13.5. Finally, we can see that the only answer choice which can give this product is m = 11 kg and d = 1.23 m.

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NGSS.MS-PS2-2

A force of 18 N is applied to a wrench that is 0.36 meters long. What is the torque applied by the wrench? Assume the force acts perpendicular to the length of the wrench.

• 18 N m

• 6.5 N m

• 4 N m

• 50 N m

• 320 N m

Answer explanation

A force of 18 N is applied to a wrench that is 0.36 meters long. What is the torque applied by the wrench? Assume the force acts perpendicular to the length of the wrench.

Torque is the product of the perpendicular force and the distance from the point of application of the force to the center of rotation.
τ = F×d = (18.0 N)(0.360 m) = 6.48 N m

A steel cable has a diameter of 2.4 cm. The tensile strength of the steel is 4 × 10⁸ N/m². What is the approximate tension force that will snap the cable?

• 4.0×10⁸ N

• 1.8×10⁵ N

• 7.5×10⁸ N

• 1.8×10¹¹ N

• 1.8×10⁹ N

Answer explanation

A steel cable has a diameter of 2.4 cm. The tensile strength of the steel is 4 × 10⁸ N/m². What is the approximate tension force that will snap the cable?

The cable will break when the stress σ reaches the tensile strength of steel, or
     σ = 4 × 10⁸ N/m².
Substituting σ = F/A tells us the approximate breaking point occurs when:
     F = σA = (4 × 10⁸ N/m²) × (4.52×10^-4 m²) = 1.81×10⁵ N

Tags

NGSS.HS-PS2-1

Which of the following expressions most nearly equals 3.3 lb ft of torque in SI units? (Use the following conversion factors: 1 ft = 0.30 m, and 1 lb = 4.45 N.)

• 3.3 N m

• 0.22 N m

• 2.5 N m

• 49 N m

• 4.4 N m

Answer explanation

Which of the following expressions most nearly equals 3.3 lb ft of torque in SI units? (Use the following conversion factors: 1 ft = 0.30 m, and 1 lb = 4.45 N.)

3.3 lb⋅ft×0.30 m1 ft×4.45 N1 lb=4.41 N m