A 5-kg crate is resting on a horizontal plank. The coeffi-
cient of static friction is 0.5 and the coefficient of kinetic
friction is 0.4. After one end of the plank is raised so
the plank makes an angle of 25
�with the horizontal, the
force of friction is?
oten
To find the force of friction acting on the crate, we can use the equation:
Force of friction = coefficient of friction * normal force
The normal force is the force exerted by the plank on the crate perpendicular to the surface. In this case, since the crate is resting on a horizontal plank, the normal force is equal to the weight of the crate.
Weight = mass * gravitational acceleration
Weight = 5 kg * 9.8 m/s² = 49 N
Now, let's calculate the force of static friction and the force of kinetic friction.
Force of static friction = coefficient of static friction * normal force
Force of static friction = 0.5 * 49 N = 24.5 N
The force of static friction acts when the crate is at rest and the applied force is not enough to overcome it.
However, once the plank is raised and the crate starts to slide, the force of static friction transitions into the force of kinetic friction.
Force of kinetic friction = coefficient of kinetic friction * normal force
Force of kinetic friction = 0.4 * 49 N = 19.6 N
Therefore, after the plank is raised, the force of friction acting on the crate is 19.6 N.
Wc = m*g = 5kg * 9.8N/kg = 49 N, = Wt.
of crate.
Fp = 49*sin25 = 20.3 = Force parallel to
ramp.
Fv = 49*cos25 = 44.4 N. = Force perpendicular to ramp.
Fs = u*Fv = 0.5*44.4 = 22,2 N. = Force
of static friction.