A block weighing 71.5 N rests on a plane inclined at 24.1° to the horizontal. The coefficient of the static and kinetic frictions are 0.26 and 0.13 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?
Wt. of block = 71.5 N.
Fp = 71.5*sin24.1 = 29.20 N. = Force
parallel to the plane.
Fn = 71.5*cos24.1 = 65.27 N. = Normal
force = Force perpendicular to the plane.
Fs = u*Fn = 0.26 * 65.27 = 16.97 N. =
Force of static friction.
F-Fp-Fs = m*a = m*0 = 0
F-29.20-16.97 = 0
F = 46.17 N.
To find the minimum magnitude of the force F that will prevent the block from slipping, we need to consider the forces acting on the block and the condition for equilibrium.
First, let's resolve the weight of the block into components parallel and perpendicular to the plane. The component of the weight parallel to the plane is given by:
F_parallel = mg * sin(θ)
where m is the mass of the block (found using the equation m = F_weight / g) and θ is the angle of inclination of the plane (24.1°).
Next, let's calculate the maximum static friction force (F_static_max) using the formula:
F_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force exerted on the block. The normal force N is equal to the component of the weight perpendicular to the plane:
N = mg * cos(θ)
Now, we can determine the maximum static friction force:
F_static_max = μ_static * mg * cos(θ)
Since the block is at rest, the force F parallel to the plane needs to match or exceed the maximum static friction force:
F_parallel ≥ F_static_max
In other words:
F_parallel ≥ μ_static * mg * cos(θ)
By substituting the known values into the equation, we can find the minimum magnitude of the force F that will prevent the block from slipping.