The mine car and its contents have a combined mass of 5 Mg and a center of gravi

Question

The mine car and its contents have a combined mass of 5 Mg and a center of gravity at G. When the wheels are locked the coefficient of static friction between the wheels and tracks is mu s, = 0.3. Work in 2 dimensions but assume that two wheels are mounted to each axle (ie, 4 wheels total) and that the symbolic parameters take the values specified below. Find the normal force acting on the front and rear wheels, assuming both are locked. Determine whether the car moves. F1 = 8 kNs, a = 0.75 m, b = 0.5 m, c = 1.25 m, r = 0.15 m

Explanation / Answer

(a) Let reactions at A and b be Ra and Rb respectively. Take gravitational constant g= 10 m/s2.

Doing moments balance about B, we get, Ra*c + F*(a+r) = (5000*10)b.

Putting values, we get Ra*1.25 + 8000*(0.75+0.15) = 50000*0.5.

So, Ra = 14240 N = 14.24 kN

So, reaction on each rear wheel = Ra/2 = 7.12 kN

Doing forces balance in vertical direction, we get Ra + Rb = 5000*10

So, Rb = 35760 N = 35.76 kN

So, reaction on each front wheel = Rb/2 = 17.88 kN.

(b) Friction force on rear wheels = Ra = 0.3*14.24 = 4.272 kN

Friction force on front wheels = Rb = 0.3*35.76 = 10.728 kN

Total friction force = 4.272+10.728 = 15 kN

The applied force F is 8 kN which is less than the static friction force. Hence the cart will not move.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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