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A) m1v1 + m2v2= (m1+m2)V V= Common velocity, v2=0 Then, 1/2(m1+m2)V=Fr*d Fr= Resistive force d= Depth of penetration (Is that right)? B) v1f= (m1-m2)*v1/(m1+m2)+2m2*v2(m1+m2) v1f (velocity after collision) v2f=(m2-m1)*v2/(m1+m2)+2m1/(m1+m2) Then, 1/2m*v1f+1/2m*v2f= Fr*d (Is that right)?

But that is ignored in solving such problems. Please any equation for part B

But how I am going to write two equations for Inelastic collision. It is understood that all the P.E is dissipated in work done against resistive force

Homework Statement A 2.9 ton weight falling through a distance of 6.5 ft drives a 0.5 ton pile 1.5 in. into the ground A) Assuming that the weight-pile collision is completely inelastic, find the average force of resistance exerted by ground B) Assuming that the resistive force in (A) remains...

Hm but why we express Tension by Two arrows?

I think it is complete explanation. Thanks!

Suppose a pulley is lifting a mass through rope so the rope is applying force to mass in form of Tension. Is the rope also applying that force to pulley i.e Is pulley being acted upon by Tension??

I am sure that the thing is applying force less than its weight!

Ok

Is the force on 1st link = (0.5)(9.8+2.5)

The forces between 1st and 2nd would be (0.4)(9.8+2.5) applied by 2nd and the same by 1st. Am I right?

But notice weight is force something is pulled BY but it is pulling in this case

Oh I supposed the points of contacts to be links. Thanks

To elaborate it I would use Newtons 3rd Law Suppose you have a mass in your hand, you are lifting it and it is in equilibrium. I am clear with the support force that is being applied by the hand at the mass. What is the force the mass is applying to hand??

I have a question. Suppose a man is lifting a Mass. The mass of course pushes it down. Now what is the force the mass is applying to hand of man is it Weight or Force of gravity because weight acts on mass itself. Then what we call it precisely?