a crate of weight fg is pushed by a force P on a horizontal floor. (a) the coefficient of static friction is us, and P is directed at angle theta below the horizontal. show that the minimum value of P that will move the crate is given by P=usFgsectheta / 1-ustantheta (b) find the minimum value of P that can produce motion when us=0.400,Fg=100N and theta=0°,15.0°,30.0°,45.0°, and 60.0°.
Thank you in advance!
a.) P=UsFgsec(theta)/1-ustan(theta)
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Teacher
To solve this problem, we will use Newton's second law of motion:
ΣF = ma
where ΣF is the sum of the forces acting on an object, m is the mass of the object, and a is the acceleration of the object.
First, let's consider the forces acting on the crate. We have the gravitational force (Fg) acting downward and the applied force (P) acting at an angle θ below the horizontal. We also have the static frictional force (Ff) acting opposite to the applied force, preventing the crate from moving until a certain threshold is reached.
(a) To find the minimum value of P required to move the crate, we need to determine when the static frictional force is at its maximum. This occurs just before the crate starts to move, so the sum of the forces in the horizontal direction will be zero.
ΣFx = 0
The horizontal forces are given by:
Pcosθ - Ff = 0
Now, we need to express Ff in terms of the variables given. The static frictional force can be given as:
Ff = usN
where us is the coefficient of static friction, and N is the normal force acting perpendicular to the surface, which is equal to the weight of the crate (Fg).
Plugging this into our equation, we get:
Pcosθ - usFg = 0
Now, solving for P, we have:
P = (usFg) / cosθ
Since secθ is equal to 1/cosθ, we can rewrite this as:
P = (usFg) * secθ
Finally, to account for the threshold of static friction limiting the applied force, we divide both sides of the equation by (1 - ustgθ), where tgθ is equal to tanθ:
P = (usFgsecθ) / (1 - ustanθ)
(b) To find the minimum value of P when us=0.400, Fg=100N, and different values of theta, we can plug these values into the equation:
For theta = 0°:
P = (0.400 * 100 * sec(0°)) / (1 - 0.400 * tan(0°))
P = (40 * 1) / (1 - 0) = 40N
For theta = 15.0°:
P = (0.400 * 100 * sec(15.0°)) / (1 - 0.400 * tan(15.0°))
P = (40 * 1.035) / (1 - 0.277) = 46.62N
Repeat the same steps for theta = 30.0°, 45.0°, and 60.0° to find the minimum values of P in each case.