help with forces?

Question

a large cube (mass = 24 kg) is being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 2.6 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

i know the equation is P-Fn=ma

Fn is the natural forcei know that you are supposed to add the natural force some where but im not sure where.

that is as far as i could get with this problem. see, i really did try to do it. i just can't finish it. please, PLEASE help me. i really am trying!

Answer 1

I think you mean "normal" force instead of "natural" force. (It's an easy mistake.)

[Edit -- Jun's mistake was: (P−friction) doesn't equal zero (P points sideways; friction points up).]

Let's call the large mass "m1" and the small mass "m2".

There are two vertical forces on the small cube:
1. force of gravity. This is (m2)(g), downward
2. force of friction. This is upward.

In order for the cube not to slide, we want those two vertical forces to be equal. That is, we want:

friction = (m2)g

The _maximum_ possible friction is (Fn)(μ) (where "μ" is 0.71, the coefficient of friction). So in order for the friction to keep the cube from sliding, the Fn must be large enough so that this is true:

(Fn)(μ) >= (m2)g

Now, as you guessed, the net force on the large cube (m1) is this:

P - Fn

So:

P - Fn = (m1)(a)

Solve for Fn:

Fn = P - (m1)(a)

At this point we seem to be stuck...BUT notice this: Since the two blocks are stuck together, the acceleration of m1 is the SAME as the acceleration of m2. That means we can write the following equation for the m2 mass (using the same "a" as in the above equation).

Net (horizontal) force on m2
= Fn = (m2)(a)

(Notice that we don't count "P" as one of the forces on m2. The only thing that m2 feels (horizontally) is the other block pushing against it, with the normal force Fn. It "knows" nothing about the P force acting on the other block.)

Anyway, we can use this to find P in terms of Fn and the masses:

Combine the previous two equations:
P - (m1)(a) = Fn = (m2)a
or:
P = (m1+m2)a
or (since a=Fn/m2):
P = (m1+m2)Fn/m2

Solving for Fn:
Fn = P(m2)/(m1+m2)

Now go back to the inequality that had "μ" in it:

(Fn)(μ) >= (m2)g

Substitute " P(m2)/(m1+m2)" for "Fn":
(P(m2)/(m1+m2))(μ) >= (m2)g

Solve for P:
P = (g)(m1+m2) / μ

Answer 2

The force 'P' is accelerating both blocks with acceleration 'a', so it is equal to:

P = (M+m) *a...........(1)

The small cube feels two vertical forces. One force is due to gravity, which acts downward, and another due to friction which acts upward, F_f. In order for the small cube to not move vertically, these forces must be equal.

F_f = m*g...........(2)

The force due to friction is given by

F_f = μ*Fn...........(3) where μ is the friction coefficient.

The little block is being pushed along horizontally to cause acceleration 'a' so it must push back with the normal force, Fn, equal to this force:

Fn = m*a...........(4)

Now put (4) into (3), then put the result into (2) and we have:

m*g = μ*m*a → g = μ*a

Solving for 'a' and plugging into equation (1) gives:

P = (M+m)*g/μ = (24+2.6)*9.8/.71 ≈ 367 Newtons.

Hope that helps.

Answer 3

Friction=CoefficientF x Fn
since horizontal force=P and for the cube to be stationary,
P-Friction=0
since there are two cubes, coefficientF x Sum of Fn=Friction