In figure the bar is uniform and weighing 500N. How large must W be if T1, and T2, are to be equal
Answers
Answer:
If the point at which the force is applied is ... 12-32, a uniform beam of weight 500 N and length 3.0 m is suspended ... sinθ = W/2T = ( 500 N)/(2 · 1200 N) = 0.2083 ⇒ θ = 12.02◦.
Explanation:
the horizontal position using a wedge at B. if the coefficient of static friction μs = ... friction between the belt and the cylinder; the tensions T1 and T2 are not equal. ... Thus, T2 must.
Answer:
1500 N
Explanation:
The bar is uniform and weighing 500 N.
2T₁ = 500 + W ----- (1)
==> (500 * 0.2 L) + (W * 0.4 L) = T₁ * 0.7 L
==> 100 + 0.4 W = 0.7 T₁ ----- (2)
0.8T₁ = 200 + 0.4 W ----- (3)
Add Equation (2) and (3)
==> 100 + 0.8 T₁ = 0.7 T₁ + 200
==> 0.1 T₁ = 100
==> T₁ = 1000
Place T₁ = 1000 in (3).
==> 0.8 * 1000 = 200 + 0.4 W
==> 800 = 200 + 0.4 W
==> 600 = .4W
==> W = 1500 N.
Hence, Weight is 1500 N.