Two blocks of mass m 1 = 15 kg and m 2 = 4 kg are sliding down an inclined plane

Question

Two blocks of mass m1 = 15 kg and m2 = 4 kg are sliding down an inclined plane (see figure below).

Two blocks of mass m1 = 15 kg and m2 = 4 kg are sliding down an inclined plane (see figure below). If the plane is frictionless, what is the normal force between the two masses? If the coefficients of kinetic friction between m1 and the plane is ?1 = 0.13, and between m2 and the plane is ?2 = 0.13, what is the normal force between the two masses? If ?1 = 0.03 and ?2 = 0.13, what is the normal force?

Explanation / Answer

(a)If the plane is friction less, the normal force(Fn) acting between the masses will be the difference of their respective weight components (gravity ffect on mass) parallel to the incline.
So Fn = (m1*g*sin ?) - (m2*g*sin ?)

= 9.81 m/s2 * sin 35degree(15 - 4) = 61.89 N

b) With friction involved, it will act against the downward force (due to gravity) on each mass; since frictional resistance is given as mu * (component of weight perpendicular to surface), the equation in part (a) will become as below:

Fn = [ (m1*g*sin ?) - (m1*g*cos?*mu1) ] - [ (m2*g*sin ?) - (m2*g*cos?*mu2) ]

Thus, Fn = g*sin? (m1 - m2) + [ g*cos? (m2*mu2 - m1*mu1) ] --------(1)

fn=9.81*sin 35(15-4) +[ 9.81*cos35(4 *0.13 -15*0.13)]

Fn = 50.39 N

c) Substituting new values of mu1 and mu2 in equation (1) above, we get

Fn = 62.45 N, same as part A; This is because in this case the frictional forces acting on the masses become equal and hence cancel each other's effect on the normal force (Fn) between them

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.