Answer to Question #173004 in Mechanical Engineering for Nana

Question

1.. A vector B, when added to the vector C = 3.0i+4.0j, yields a resultant vector that is in the positive y-direction and has a magnitude equal to that of C. What is the magnitude of B?

      

.2   a. Will a body moving in a circular track have zero acceleration?  Explain your answer.     

                                   

b. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. If the height of the counter is 0.860m, (a) with what speed did the mug leave the counter, and (b) what was the direction of the mug’s velocity just before it hit the floor? 

Answer 1

1) Given as,

The vector C = 3.0i+4.0j hence magnitude of vector C=$\sqrt{(3)^2+(4)^2}=5 units$

Let vector B=Ui+V j ,

If vector B is added to the vector C the resultant vector R=B+C =(U+3)i+(V+4)j.

Since resultant vector is in the Y-direction, The X-component of vector R is zero.

Therefore, U+3=0 => U=-3

Also given as magnitude o resultant vector is equal to that of vector C,

Therefore, magnitude of R= V+4=5, => V=1

Therefore magnitude of resultant vector R=$\sqrt{U^2+V^2}=\sqrt {(-3)^2+(1)^2}=\sqrt{10}$ Units

yields a resultant vector that is in the positive y-direction and has a magnitude equal to that of C. What is the magnitude of B?

2)

a) When a body moves in a circular track, it's acceleration cam never be zero. There are two acceleration acting on the body moving in circular path, tangential and radial acceleration. If body moves with constant Linear speed, it's tangential acceleration becomes zero but radial acceleration exists due to change in the direction of body along the circular path, hence there is always acceleration present in the body moving along curved path.

b)Given as, H=0.86 m, L=1.4 m

The vertical velocity at landing point-2:

$V_{2y}=\sqrt{2×g×H}=\sqrt {2×9.8×0.86}= 4.106 m/s$

The time (t) taken by mug to move from point 1 to point 2 is :

$Time(t)=\sqrt{\frac{2×H}{g}} =\sqrt{\frac{2×0 .86}{9.8}}=0.419 seconds$

The horizontal velocity at landing point-2 is:

$V_{2x}= \frac {Horizontal distance(L)}{Time(t)}=\frac{1.4}{0.419}=3.342 m/s$

(a) Velocity of mug leaving the counter (V1x)=V2x=3.342 m/s

(b) Direction of mug velocity at landing point-2 :$\theta =tan^{-1}[\frac{V_{2y}}{V_{2x}}]=50.86 \space degrees \space with \space horizontal ( as \space shown \space in \space figure)$