Category Archives: Engineering Mathematics Blog

Grantham PHY220 Week 2 Assignment Problem 7

A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47^{\circ}  angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box?

SOLUTION:

The free-body diagram of the box

week 2 problem 7

Solve for the normal force

\sum F_y=ma_y

F_N-mg+F_psin\left(47.0^{\circ} \right)=0

F_N-7.93\left(9.80\right)+84.0\:sin\left(47^{\circ} \right)=0

F_N=16.28\:N

Solve for the friction force

F_{fr}=\mu _kF_N=0.35\left(16.28\:N\right)=5.70\:N

Solve for the acceleration in the horizontal direction

\sum F_x=ma_x

F_p\:cos\:\left(47^{\circ} \right)-F_{fr}=7.93\left(a_x\right)

84.0\:N\cdot cos\:\left(47^{\circ} \right)-5.70\:N=7.93\:kg\:\cdot \left(a_x\right)

a_x=\frac{51.59\:N}{7.93\:kg}=6.51\:m/s^2

Therefore, the acceleration of the box is 6.51 m/s2 along the horizontal surface.

Grantham PHY220 Week 2 Assignment Problem 6

If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction?

SOLUTION:

The acceleration due to gravity on the moon is

g_m=\frac{1}{6}\left(9.80\:m/s^2\right)=1.63\:m/s^2

The weight of a 100-kg man on the moon is

W_m=mg_m=\left(100\:kg\right)\left(1.63\:m/s^2\right)=163.3\:N

If the elevator is accelerating upward then the acceleration would be greater. The person would be pushed toward the floor of the elevator making the weight increase. Therefore, the elevator must be going down to decrease the acceleration.

For a 100 kg man to experience a 163.3 N in an elevator,

F=ma

163.3\:N=100\:kg\:\left(9.80\:m/s^2-a_e\right)

9.80-a_e=\frac{163.3}{100}

a_e=9.80-\frac{163.3}{100}

a_e=8.167\:m/s^2

Therefore, the elevator should be accelerated at 8.167 m/s2 downward for a 100-kg man to simulate his weight just like his weight in the moon which has 1/6 of the Earth’s gravity acceleration.

Grantham PHY220 Week 2 Assignment Problem 4

A not so brilliant physics student wants to jump from a 3rd-floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make it to the pool’s edge?

Solution:

Since the student will be running horizontally, there is no initial vertical velocity, v_{0_y}=0. We are also given \Delta x=8\:m, and \Delta y=-20\:m.

Consider the vertical component of the motion.

\Delta y=v_{0_y}t-\frac{1}{2}gt^2

-20=0-\frac{1}{2}\left(9.80\right)t^2

-20=-4.9t^2

t^2=\frac{20}{4.9}

t=\sqrt{\frac{20}{4.9}}

t=2.02\:s

Consider the horizontal component of the motion

\Delta x=v_{0_x}t

v_{0_x}=\frac{\Delta x}{t}

v_{0_x}=\frac{8}{2.02}

v_{0_x}=3.96\:m/s

Therefore, the student should be running 3.96 m/s horizontally to make it to the pool’s edge.

Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

Grantham PHY220 Week 2 Problem 2

An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.

Solution:

The angle of the plane in the x-y axis

Solve for \theta

\theta =tan^{-1}\left(\frac{1500}{2100}\right)=35.5^{\circ}

The angle is 35.5 degrees measured from the west axis. If it is measured from the East axis, the angle would be 180-35.5=144.5 degrees.

The distance from the plane to the observer

Solve for d=\sqrt{\left(1500\:m\right)^2+\left(2100\:m\right)^2+\left(5000\:m\right)^2}=5626.72\:m

Grantham PHY220 Week 2 Assignment Problem 1

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power?

Solution:

Week 2 Problem 1

R_x=-3\:cos\:55^{\circ }=-1.720729309\:m/s

R_y=3\:sin\:55^{\circ }-2=0.4574561329\:\:m/s

The x component of the resultant is negative and the y component is positive, thus the resultant is located at the second quadrant.

R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-1.720729309\:m/s\right)^2+\left(0.4574561329\right)^2}=1.78\:m/s

\theta =tan^{-1}\left(\frac{R_y}{R_x}\right)=tan^{-1}\left(\frac{0.4574461329}{-1.720129309}\right)=-14.9^{\circ}

Therefore, the magnitude of the net velocity of the ship is 1.78 m/s, and is going 14.9 degrees North of West

Grantham PHY220 Week 2 Assignment

Problems:

  1. A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power? [Solution]

  2. An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y-axis and the total distance to the plane from you. Choose the x-axis east, y-axis north, and z-axis up. [Solution]

  3. A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s? [Solution]

  4. A not so brilliant physics student wants to jump from a 3rd-floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make it to the pool’s edge? [Solution]

  5. If a 1500 kg car stopped from an in 5.6 seconds with an applied force of 5000 N, how fast was it initially traveling?[Solution]

  6. If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction? [Solution]

  7. A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47.0° angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box? [Solution]

  8. If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road? [Solution]

 

Theory of Failure List of Questions with Answers

  1. σ_1 is the only stress to use to find the factor of safety by the maximum normal stress theory of failure
    • True
    • False
  2. A load is changing on a part from 1,000 lb to 5,000 lb, the average and range loads are
    • 6,000 lb and 4,000 lb
    • 5,000 lb and 1,000 lb
    • 3,000 lb and 2,000 lb
    • 1,000 lb and 5,000 lb
  3. A part is loaded under service stress for 25,000 reversals, then under a higher stress for 35,000 reversals until failure. If the total life under the larger stress alone is 36,000 reversals, what is the estimated life under the service load?
    • 900,000 reversals
    • 37,000 reversals
    • 50,000 reversals
    • 60,000 reversals
  4. Brittle materials fail in yielding
    • True
    • False
  5. By applying 10,000 in.lb torque to this non-rotating shaft, the torque reactions are
    • T1=5,000 in.lb; T2=5,000 in.lb
    • T1=6,000 in.lb; T2=4,000 in.lb
    • T1=4,000 in.lb; T2=6,000 in.lb
    • None of the above
  6. For a ductile material, the yield stress in shear is the same as it is for tension
    • True
    • False
  7. For a loaded part, the state of stress is σ_x=2,000 psi, σ_y=-8000 psi, τ_xy=0. If σ_yp=45,000 psi, what is the factor of safety using the maximum normal stress theory of failure?
    • 5.625
    • 22.5
    • 4.5
    • 7.5
  8. For a safe design in an alternating loading situation by the modified Goodman approach, two equations must be satisfied
    • True
    • False
  9. For a shaft that is transmitting 145 HP while turning at 1,200 rpm, the torque is
    • 760.2 in.lb
    • 6722 in.lb
    • 7615.5 in.lb
    • 2235.3 in.lb
  10. For a shaft under torque, the only type of deformation that will result is the rotation of the cross-sections with respect to each other
    • True
    • False
  11. For a brittle material, there are two curves for the stress-strain relationship–one for the tension and the other for compression.
    • True
    • False
  12. If a part fails by yielding, it will sustain permanent deformation
    • True
    • False
  13. In order to apply one of the failure theories to estimate the factor of safety, we have to
    • use the state of stress
    • find the principal stresses
    • calculate the maximum shear stress
    • find the yielding stress of the material
  14. Maximum shear theory of failure only applies to brittle materials
    • True
    • False
  15. Stress concentration factors are functions of
    • Load
    • Geometry
    • Material
    • Geometry and type of loading
  16. Stress concentration factors must be applied when we are using brittle materials
    • True
    • False
  17. The principal stresses for a part are σ_1=20 MPa, σ_2=5 MPa. What is the factor of safety using the Mises Henckey theory of failure if σ_yp=100 MPa?
    • 6.1
    • 6.5
    • 5.55
    • 8.23
  18. The shear stress on a cross-section of a shaft under torsion is linearly proportional to its radius
    • True
    • False
  19. A torque applied to one end of a shaft, while the other end is keyed against rotation. The torque along the shaft will be proportional to the length
    • True
    • False
  20. Ultimate stress is the same as yielding stress
    • True
    • False
  21. When solving for the factor of safety in a combined steady and alternating loading situation, stress concentration factors must be applied for both average and range stresses.
    • True
    • False

Grantham PHY220 Week 1 Assignment Problem 8

Problem 8

A stone is dropped from the roof of a high building. A second stone is dropped 1.25 s later. How long does it take for the stones to be 25.0 meters apart?

Solution:

Let t be the amount of time after the first stone is dropped. The distance from traveled by the first stone is

y_1=\frac{1}{2}gt^2

The distance traveled by the second stone is

y_2=\frac{1}{2}g\left(t-1.25\right)^2

The difference between the two stones is 25.0 m after time

y_1-y_2=25.0

\frac{1}{2}\left(9.80\:m/s^2\right)t^2-\frac{1}{2}\left(9.80\:m/s^2\right)\left(t-1.25\right)^2=25

4.9t^2-4.9\left(t^2-2.5t+1.5625\right)=25.0

4.9t^2-4.9t^2+12.25t-7.65625=25.0

12.25t=25.0+7.65625

12.25t=32.65625

t=2.67\:s