Suppose there is friction (static coefficient = 0.22, and kinetic coefficient = 0.15). With this friction, what force (if any) is necessary to hold the 30kg mass stationary on a 35 degree incline?
Wm = m*g = 30kg * 9.8N/kg = 294 N. = Wt.
of the mass.
Fp = 294*sin35 = 168.6 N. = Force parallel to the incline.
Fn = 294*cos35 = 240.8 N. = Normal force
or force perpendicular to the incline.
Fs = u*Fn = 0.22 * 240.8 = 52.98 N. =
Force of static friction.
Fap-Fp-Fs = 0
Fap = Fp+Fs = 168.6 + 52.98 = 221.6 N. =
Force applied.
To determine the force necessary to hold the 30kg mass stationary on a 35-degree incline with friction, you'll need to consider the gravitational force pulling the object down the incline and the frictional force opposing its motion.
First, find the gravitational force acting on the object. The formula for gravitational force is F = m * g, where "m" is the mass of the object and "g" is the acceleration due to gravity (approximately 9.8 m/s²).
F_gravity = 30 kg * 9.8 m/s²
F_gravity = 294 N
Next, calculate the force of friction. The force of friction can be determined using the equation F_friction = μ * N, where "μ" is the coefficient of friction and "N" is the normal force (the force perpendicular to the incline).
The normal force can be calculated by multiplying the weight of the object by the cosine of the angle of inclination.
N = m * g * cos(θ)
N = 30 kg * 9.8 m/s² * cos(35°)
Now, calculate the force of friction using the given coefficients of friction.
F_friction = μ * N
F_friction = (0.22) * [30 kg * 9.8 m/s² * cos(35°)]
Finally, the force necessary to hold the mass stationary is the force of friction in the opposite direction of the gravitational force, so the net force is:
F_net = F_gravity - F_friction
Substituting the previously calculated values into the equation will give you the final answer.