Hibbeler Engineering Mechanics, 11th Ed. Chapter 15, Question 36 Two men A and B

Question

Hibbeler
Engineering Mechanics, 11th Ed.
Chapter 15, Question 36
Two men A andB, each having a weight of 160 lb,stand on the 200 lb cart. Each runs with a speed of 3ft/s measured relative to the cart. Determine the final speedof the cart if
(a) A runs and jumps off, and thenB begins to run and jump off of the sameend, and
(b) both run and jump off the cart at the same time, on the sameend.
Assuming jumps are made horizontally and neglect the mass of thewheels.

The answer key they have for this is brutally done. No use tome as I don`t understand what it`s trying to say. (Unexplained variables)   Any clarification would begreatly appreciated.

Explanation / Answer

Given that Weight of men are 160lb and Weight of cart as200lb The Speed of men relative to cart isV= 3ft/s a)Initially the cart and the men are at rest so intialmomentum is zero. Then the cart and men move relative to each other. Then A runs and jumps off the cart so now the final momentumbecomes the momentum of the man when he runs along with the cartand the momentum of the cart and man B when the man A jumpsoff. The Velocity of the man is the relative velocity with thecart.Lets assume final velocity of cart=V1 As the man and cart are moving in same direction the velocityof man is (V1-V) When A jumps off the cart the total mass will be mass of cart+mass of man B Now Applying Conservation of momentum 0=Wm*(V1-V)+(Wm+Wc)V1 From this equationV1=WmV/(2Wm+Wc) =160*3/(520) =0.923ft/s
Now B begins to run and jump offthe Cart. Now the cart is moving with A onit with V1 speed .so initial momentum will be(Wm+Wc)V1 Now Lets think the final velocityof cart will be V2 when B starts running. Now when initially B runs hismomentum will be Wm(V2-V) Now the Momentum of cart after Bjumps off Will be WcV2 Now Applying Conservation ofmomentum (Wm+Wc)V1=Wm(V2-V)+WcV2 Now FromAbove EquationV2=((Wm+Wc)V1+WmV)/(Wm+Wc) =((160+200)0.923+160*3)/(160+200) =2.256ft/s b)Now in the second part A and Bjump off Together at a time As we know the initial momentumwould be zero .Now let V3 be the final velocity ofcart. Now when A anb B run and jump off ata time the momentum of A with respect to cart will be Wm(V3-V)
similarly the Momentum of B whenRuns is Wm(V3-V) as they both run relative to cart. And the final momentum of cart whenboth the men jump off the cart will be WcV3 so Now applying conservation ofmomentum for this condition 0=Wm(V3-V)+Wm(V3-V)+WcV3 Now V3=2WmV/(2Wm+Wc) =2*160*3/(2*160+200)
=1.846ft/s
Hope you understand thisexplanation.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.