(a) Show that the general expression for the horizontal force P required to push

Question

(a) Show that the general expression for the horizontal force P required to push a block of weight w at constant velocity up the sloping surface of an inclined plane at angle theta above the horizontal, when the coefficient of sliding friction is u, is P=w(sin(theta)+ucos(theta)/cos(theta)-usin(theta)).

you should be able to understand the problem and draw a free-body diagram from the statement of the problem.

(b) Suppose that w = 100lb and u=0.75. Compute values of P for angles: theta=0, 37, 45, 53, 60. Sketch a graph of P vs. theta

(c) from the general expression for P given in (a), find angle theta for which an infinite force is required.

(d) When theta is greater than this angle, what has to be true about P (and N) for equilibrium ?

Explanation / Answer

There are four forces acting on the box:
1. Weight (gravity);
2. Normal force (call it Fn);
3. The diagonal force (call it F)
4. Friction (call it Fr. Note that, by the law of static friction, Fr ? ?Fn)

List the horizontal and vertical components of all forces acting on the box. Note that since the box is not moving, the forces are balanced (sum to zero) in both the horizontal and vertical directions.

First the vertical components (let "up" be positive)
1. Vertical component of weight: ?mg
2. Vertical component of normal force: +Fn
3. Vertical component of "F": ?Fsin?
4. Vertical component of friction: zero

Vertical components sum to zero, so:
?mg + Fn ? Fsin? + 0 = 0

Now the horizontal components:
1. Horizontal component of weight: zero
2. Horizontal component normal force: zero
3. Horizontal component of "F": +Fcos?
4. Horizontal component of friction: ?Fr

Horizontal components sum to zero, so:
Fcos? ? Fr = 0

The rest is just algebra. From the equation above:
Fr = Fcos?

Combine that with the inequality, Fr??Fn:
Fcos? ? ?Fn

Combine with the first equation (vertical components) to eliminate Fn:
Fcos? ? ?(mg + Fsin?)

Divide by ?Fcos? and rearrange terms:
tan? ? 1/? ? mg/(Fcos?)

That means: if tan? is greater than or equal to the given amount, the box won't move. Note that if tan? is 1/? or greater,

normal force should be less than mg because you're giving some upward force--
n=mg-Fxsin(theta)-- u=f/n should give the right answer

Edit:
First, you know how much work you do on the sled-- W=F*x-- (215N)(cos(35))*6.9m=1215.2 J

Next, you know that friction does uN*x of work on the sled-- and the difference between your work and the friction's should be the net work--220J
so the work done by friction is 995.2 J

the frictional force is W(friction)/x=995.2J/6.9m =144.2 N.

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