From my understanding the following should be correct?
F(normal) = m1*g therefore F(friction) = μ(k) * m1 * g
ƩF = F(applied) - F(friction) = (m1 + m2)*a
F = (m1+m2)*(g/μ(s)) + μ(k)*m1*g
Although this gives me a very large applied force...
F(friction) isn't μ(k) * (m1 + m2) * g is it?
@HallsofIvy .8(9.8) > .47(.8(F/9.4)) therefore (if I did the calculations correctly) 196 > F How does this help me solve for F ? I am probably misunderstanding your point...
@azizlwl Wouldn't the force needed to hold up the block be F(net on small block) = F - (.47*m2*a) and F(net) = 0 ...
Homework Statement
The static friction between the two blocks is .47 and kinetic friction between the block with mass 9.4 kg and the horizontal surface is 0.21.The acceleration of gravity is 9.8 m/s^2 What is the minimum force F which must be exerted on the 9.4 kg block in order that the 0.8...