A rectangular block of mass 1 kg is trapped below a wedge (mass=10 kg). The dimensions of this block are negligible compared to those specified in the diagram.
One plans to take the block out by applying a pulling force to the left.
Answer the following questions:
1)What minimum force must be applied to accomplish this?
2)If this bare minimum force is applied:
a)What is the initial acceleration?
b)If the same force is applied to the right (pushing), what is the maximum displacement of the block?
c)What is the accn. when the block starts to come out of the bottom of the wedge?
d)If one continues to apply the same force for a very long time, what is
the acceleration at t=inf.
Assume the coeff. of friction at all contacts= .5.
isn't the vertical dimension of the wedge given?
ReplyDeleteFOR FIRST PART FORCE = 70.2333 N
ReplyDeleteIS MY ANSWER CORRECT BHAIYA?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteFmin 43/6 g
ReplyDeleteinitial accl = 0
when it comes out accl = 10/3 g
at infinity 20g/3
1.43g/6
ReplyDelete2.
(a)0
(b)displacement=0
(c)8g/3
(d)20g/3
bhaiya please reply if it is correct
ReplyDelete@amit and ankush..: shudn't the last answer be infinity? a finite accelaration acting for infinite time...?? hw did u do it?
ReplyDeleteby the way, for third part, i got sqrt(80g(1-ln2))
the previous comment ws by me.. sorry for the anonymous tag.!
ReplyDelete