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The Engineering Crucible
United States
เข้าร่วมเมื่อ 2 ส.ค. 2018
Need help with your engineering homework? Get help from an actual engineering tutor.
This channel will cover problems frequently encountered at the university / college level in statics, dynamics and mechanics of materials.
The main problems we will be covering are the fundamental problems from the beginning of the set of problems for each section of the chapters. The reason being, is that they have the answers at the end of the book in case you want to work through them on your own, or to verify my work ;)
In the future I will aim to do problems from the actual sections, and cover problems I saw often at SDSU, USD.
Commonly referenced books by my students: Ferdinand Beer and Hibbler
Have questions about specific problems, leave a request in the comments!
This channel will cover problems frequently encountered at the university / college level in statics, dynamics and mechanics of materials.
The main problems we will be covering are the fundamental problems from the beginning of the set of problems for each section of the chapters. The reason being, is that they have the answers at the end of the book in case you want to work through them on your own, or to verify my work ;)
In the future I will aim to do problems from the actual sections, and cover problems I saw often at SDSU, USD.
Commonly referenced books by my students: Ferdinand Beer and Hibbler
Have questions about specific problems, leave a request in the comments!
Problem F14-18 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
Conservative forces and potential energy.
The 4-kg collar C has a velocity of v_a = 2 m/s when it is at A. If the guide rod is smooth, determine the speed of the collar when it is at B. The spring has an unstretched length of l_o = 0.2 m.
The 4-kg collar C has a velocity of v_a = 2 m/s when it is at A. If the guide rod is smooth, determine the speed of the collar when it is at B. The spring has an unstretched length of l_o = 0.2 m.
มุมมอง: 8 733
วีดีโอ
Problem F14-17 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
มุมมอง 4.1K3 ปีที่แล้ว
Conservative forces and potential energy. The 75-lb block is release from rest 5 ft above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched.
Problem F14-16 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
มุมมอง 7K3 ปีที่แล้ว
Conservative forces and potential energy. The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it strikes the stop B. The spring has an unstretched length of 0.5 ft.
Problem F14-15 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
มุมมอง 11K3 ปีที่แล้ว
Conservative forces and potential energy. The 2-kg collar is given a downward velocity of 4 m/s when it is at A. If the spring has an unstretched length of 1 m and a stiffness of k = 30 N/m, determine the velocity of the collar at s = 1 m.
Problem F14-14 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
มุมมอง 3.7K3 ปีที่แล้ว
Conservative forces and potential energy. The 2-kg package leaves the conveyor belt at A with a speed of v_a = 1 m/s and slides down the smooth ramp. Determine the required speed of the conveyor belt at B so that the package can be delivered without slipping on the belt. Also, find the normal reaction the curved portion of the ramp exerts on the package at B if rho_b = 2 m.
Problem F14-13 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Conservation of Energy
มุมมอง 6K3 ปีที่แล้ว
Conservative forces and potential energy. The 2-kg pendulum bob is released from rest when it is at A. Determine the speed of the bob and the tension in the cord when the bob passes through its lowest position, B.
Problem F14-9 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Power and Efficiency
มุมมอง 6K3 ปีที่แล้ว
Principal of work and energy. If the motor winds in the cable with a constant speed of v = 3 ft/s, determine the power supplied to the motor. The load weighs 100 lb and the efficiency of the motor is epsilon = 0.8. Neglect the mass of the pulleys.
Problem F14-12 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Power and Efficiency
มุมมอง 4.4K3 ปีที่แล้ว
Principal of work and energy. At the instant shown, point P on the cable has a velocity v_p = 12 m/s, which is increasing at a rate of a_p = 6 m/s^2. Determine the power input to motor M at this instant if it operates with an efficiency epsilon = 0.8. The mass of block A is 50 kg.
Problem F14-8 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Power and Efficiency
มุมมอง 4.1K3 ปีที่แล้ว
Principal of work and energy. If F = 10s N, where s is in meters, and the contact surface between the block and the ground is smooth, determine the power of force F when s = 5 m. When s = 0, the 20 kg block is moving at v = 1 m/s.
Problem F14-6 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Work and Energy
มุมมอง 5K3 ปีที่แล้ว
Principal of work and energy. The 5-lb collar is pulled by a cord that passes around a small peg at C. If the cord is subjected to a constant force F = 10 lb, and the collar is at rest when it is at A, determine its speed when it reaches B. Neglect friction.
Problem F14-5 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Work and Energy
มุมมอง 15K3 ปีที่แล้ว
Principal of work and energy. When s = 0.6 m, the spring is unstretched and the 10-kg block has a speed of 5 m/s down the smooth plane. Determine the distance s when the block stops.
Problem F14-2 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Work and Energy
มุมมอง 12K3 ปีที่แล้ว
Principal of work and energy. If the motor exerts a constant force of 300 N on the cable, determine the speed of the 20 k crate when it travels s = 10 m up the plane, starting from rest. The coefficient of kinetic friction between the crate and the plane is mu_k = 0.3.
Problem F14-1 Dynamics Hibbeler 13th (Chapter 14) Engineering Dynamics - Work and Energy
มุมมอง 16K3 ปีที่แล้ว
Principal of work and energy. The spring is placed between the wall and the 10-kg block. If the block is subjected to a force of F = 500 N, determine its velocity when s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.
Problem 13-107 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 8K3 ปีที่แล้ว
Equations of motion: Cylindrical Coordinates The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limacon, r = (2 cos(theta)) ft. If theta = 0.5t^2 rad/s, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side.
Problem 13-106 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 1.9K3 ปีที่แล้ว
Equations of motion: Cylindrical Coordinates The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limacon, r = (2 cos(theta)) ft. If at all times theta_dot = 0.5 rad/s, determine the force which the rod exerts on the particle at the instant theta = 60 degrees. The fork and path contact the particle on only one side. Solve prob. 13-105 at the insta...
Problem 13-105 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 3.2K3 ปีที่แล้ว
Problem 13-105 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem 13-92 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 4.8K3 ปีที่แล้ว
Problem 13-92 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem 13-89 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 16K3 ปีที่แล้ว
Problem 13-89 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-15 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 5K3 ปีที่แล้ว
Problem F13-15 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-16 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 20K3 ปีที่แล้ว
Problem F13-16 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-10 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 24K3 ปีที่แล้ว
Problem F13-10 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-14 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 5K3 ปีที่แล้ว
Problem F13-14 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-11 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 12K3 ปีที่แล้ว
Problem F13-11 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
Problem F13-7 & F13-8 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
มุมมอง 11K3 ปีที่แล้ว
Problem F13-7 & F13-8 Dynamics Hibbeler 13th (Chapter 13) Engineering Dynamics
11-50 Vector Mechanics for Engineers Statics|Dynamics C11 (10th Edition)
มุมมอง 4.9K3 ปีที่แล้ว
11-50 Vector Mechanics for Engineers Statics|Dynamics C11 (10th Edition)
Problem F13-6 Dynamics Hibbeler 13th (Chapter 13)
มุมมอง 17K3 ปีที่แล้ว
Problem F13-6 Dynamics Hibbeler 13th (Chapter 13)
Problem F13-4 Dynamics Hibbeler 13th (Chapter 13)
มุมมอง 11K3 ปีที่แล้ว
Problem F13-4 Dynamics Hibbeler 13th (Chapter 13)
Problem F13-5 Dynamics Hibbeler 13th (Chapter 13)
มุมมอง 13K3 ปีที่แล้ว
Problem F13-5 Dynamics Hibbeler 13th (Chapter 13)
Problem F13-3 Dynamics Hibbeler 13th (Chapter 13)
มุมมอง 22K3 ปีที่แล้ว
Problem F13-3 Dynamics Hibbeler 13th (Chapter 13)
Problem F13-2 Dynamics Hibbeler 13th (Chapter 13)
มุมมอง 17K3 ปีที่แล้ว
Problem F13-2 Dynamics Hibbeler 13th (Chapter 13)
i dont get it doesnt that car want to slip downwards? so shouldnt the friction be going up?
where did the 32.2 come from?
gravity
Hello, why was the value of theta substituted in Fsin theta and not F=60.6/cos theta?
you cant even talk i cant stand with 2x, not enough
Is this the same method of force and acceleration method?
Thank youuuuuuuu
What is the reasoning for using the 60 degrees instead of 30 degrees for the force of 250 N?
Why is that summation of moments about A is Cy(6) -450(3.46)=0 Why its -450 and not positive 450 when the arrow is pointing to the right, and its positive
Finally found someone who explains it easi
good day sir! i would like to ask why did you not decompose the weight into it's x and y components? is there no such need?
broooo thank you so much for you work you are helping me alot during my studying god bless your efforts and please continue this
How is Ax a positive value when it's facing in the negative direction?
What if we take the datum down
Why doesn't the 600N force create a moment around the x-axis??
God of war analogy is so real
im really having fun watching every video and learning about statics at the same time 😂 i love your goofiness haha
Useful insights, nice method. Thanks
Helpful explanation. Many thanks!
Glad it helped!
Thanks for the great FBD and explanation
Great FBDs. Thanks very much for your clear explanations.
Hello, I was wondering if you, by any chance, could cover chapter 3 of hibbeler?
shouldnt we take in to accoun the u axis? what are they trying to achieve by placing the U axis??
hello, i have a question i tried using the left side but my answer was (12-4(6x))
Thank you!
Thanks
Helpful, thanks
Hello I have a question, how do you know which direction u are to put positive or negative on ?
its assumed that up and to the right is always positive.
Even arabs appreciates your work, shukran❤
Oh great heavens, what is going on here
감사합니다
감사합니다
Thanks
the algebra doesnt make sense for teh quadratic equation s^2: -100 s: 100-49.05 = 50.95 constant: 125-60+36+29.43 = 130.43 tell me if im wrong please
Hi mr, i need your help to teach me dynamics
Its + cos(30) not -sin(30) because its on a decline so when we resolve the wight its going to be to the right and down we need the force to the right cause its gonna push with the force so +10x9.81cos(30)(S-0.6) the rest is correct
Incorrect, gravity acts in the Y direction, so we need to use sin to find the force in the Y.
From India
Thanks
⭐️⭐️⭐️
I love this guy
its hard for me to follow if you fill in the variables at the end. It resolves in a long sentence of variables wich makes he dizzy in a kind of way, haha. maybe you can fill in the variables sooner. that way you dont get the T = m*ax+m*g*sin30+uk*mg*cos 30. I have to read it ten times to understand (13:37) But above all, thanssk for the explanation :)
Try to get used to entering numbers at the end. This will help you out in the long run. Many terms will simplify/cancel out in your future physics/engineering classes
I agree
Dude tomorrow is my final and I’ve been watching your videos since the semester started and it helped me a lot so thank you so much and may god bless you ❤️❤️
tnx man
❤❤❤❤❤❤❤❤❤❤❤❤❤ thank you so much
Nice job with the derivation, as i could follow along. Instead of integrating, could you have substituted acceleration using the kinematic eqn for acceleration V^2=V0^2+2as?
Can someone explain why weight isn't accounted for in this problem?
We are solving for forces in the x-plane (horizontal plane) as the problem states. Weight is only acting in the vertical/y-plane. Hope that clears it up!
I thought since the plane the block is sliding is smooth therefore, there is no friction. So then why are we figuring out T_1 if it is supposed to be 0?
T1 no puede ser cero porque ya viene con una velocidad, T2 es 0 porque frena y no existe movimiento, sorry 6 month later
thank you for tis video, very helpful!
Why is the at= 6?
Tangential acceleration is the acceleration that causes change in speed
besttt teacherr
Thank youuuuuu you are the best