(a) What is the minimum force of friction required to hold the system of Figure

Question

 

 

(a) What is the minimum force of friction required to hold the system of Figure P4.58 in equilibrium? Let w1 = 120 N and w2 = 35.0 N.
N
(b) What coefficient of static friction between the 120 N block and the table ensures equilibrium?

(c) If the coefficient of kinetic friction between the 120 N block and the table is 0.175, what hanging weight should replace the 35.0 N weight to allow the system to move at a constant speed once it is set in motion?
N

(a) What is the minimum force of friction required to hold the system of Figure P4.58 in equilibrium? Let w1 = 120 N and w2 = 35.0 N. N (b) What coefficient of static friction between the 120 N block and the table ensures equilibrium? (c) If the coefficient of kinetic friction between the 120 N block and the table is 0.175, what hanging weight should replace the 35.0 N weight to allow the system to move at a constant speed once it is set in motion? N

Explanation / Answer

(a) tension in the string T = w 2 = 35N forblock of weight w 1 , Tension is along +vedirection and frictional force is along -vedirection.If tension is equal to frictional force the system is inequilibrium. So ,required frictional force f = 35 N ( b ) .we know frictional force f = N where N =normal force = w 1 = 120 N fromthis we find coefficient of static friction between the120 N block and the table ensures equilibrium i.e., = 35.0/120 = 0.291 (c) given k = 0.175 the force of friction, f = 0.174*120 = 20.88 N if the system moving constant speed, the net force is zero, ==> T - f = 0 T = f = 20.88 N the weight acting on the black is 20.88 N.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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