A uniform beam of mass M is supported in a horizontal position by a pin and cabl

Question

A uniform beam of mass M is supported in a horizontal position by a pin and cable as shown in the Figure. Given. M = 32.0 kg ; m = 69.0 kg ; x = 2.50 m ; L = 5.10 m ; and ? = 27.0

A uniform beam of mass M is supported in a horizontal position by a pin and cable as shown in the Figure. Given. M = 32.0 kg ; m = 69.0 kg ; x = 2.50 m ; L = 5.10 m ; and ? = 27.0 A degree. a) How many forces are acting on the beam (treat the interaction at the pivot with the wall as 2 forces)? b) What is in N m, the torque due to the mass of the beam? (Use counterclockwise as positive) c) What is in N the tension of the cable? d) What is in N, H, the horizontal component of the force exerted by the wall onto the beam? Assume the positive direction points towards the right. e) What is in N, V, the vertical component of the force exerted by the wall onto the beam? Assume the positive direction points upwards. Use 10.0 N/kg for g.

Explanation / Answer

a) 4

b) Torque Q1 = M*g*(L/2) = 32*9.8*(5.1/2) = 816 Nm

c) net torque = 0

N*sin27*x - Q1 - m*g*L = 0

N*sin27*2.5 = 816 +(69*10*5.1) = 4335

N = 3819.46 N

d)

H + N*cos27 = 0

H = -N*cos27 = -3403.17 N

e) V + N*sin27 - M*g - mg = 0

V = -723.99 = 724 N --------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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