Help me confirm if this is correct and drawing a diagram please (resolution attached) A 15,000 N crane can rotate around a frictionless shaft at its base and is supported by a cable that forms a 25-degree angle with the crane (see figure). The crane has 16 m long and not uniform; its center of gravity is 7.0 m from the shaft measured along the crane. The cable is subject to 3.0 m from the top end of the crane. The crane rises to 55o above the horizontal, holding a container with 11,000 N bricks by a very light rope of 2.2 m. Calculate a) the voltage on the cable and b) the vertical and horizontal components of the force exerted by the shaft on the crane. Start by drawing a free body diagram of the crane.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Help me confirm if this is correct and drawing a diagram please (resolution attached)

A 15,000 N crane can rotate around a frictionless shaft at its base and is supported by a cable that forms a 25-degree angle with the crane (see figure). The crane has
16 m long and not uniform; its center of gravity is 7.0 m from the shaft measured along the crane. The cable is subject to 3.0 m from the top end of the crane. The crane rises to
55o above the horizontal, holding a container with 11,000 N bricks by a very light rope of 2.2 m. Calculate a) the voltage on the cable and b) the vertical and horizontal components of the force exerted by the shaft on the crane. Start by drawing a free body diagram of the crane.

Datos:
Peso de la grúa: w, = 15000 N
Longitud de la grúa: L, = 16m
Peso de los ladrillos: W; = 11000 N
a. Ahora, si el sistema se encuentra en equilibrio, el momento neto en cualquier punto es
cero, donde T es la tensión del cable:
T sen25(16 – 3) = W, (7)(cos55) + W;(16)(cos55)
15000(7)(0.573) + 11000(16)(0.573
T =
0.422(13)
T = 29349.8 N
b. Fuerza horizontal
= Tcos30
= 29349.8(cos30)
= 25417.67 N
Fuerza vertical
= Tsen30 + 15000 + 11000
= 29349.8(0.5) + 15000 + 11000
= 40674.5 N
Transcribed Image Text:Datos: Peso de la grúa: w, = 15000 N Longitud de la grúa: L, = 16m Peso de los ladrillos: W; = 11000 N a. Ahora, si el sistema se encuentra en equilibrio, el momento neto en cualquier punto es cero, donde T es la tensión del cable: T sen25(16 – 3) = W, (7)(cos55) + W;(16)(cos55) 15000(7)(0.573) + 11000(16)(0.573 T = 0.422(13) T = 29349.8 N b. Fuerza horizontal = Tcos30 = 29349.8(cos30) = 25417.67 N Fuerza vertical = Tsen30 + 15000 + 11000 = 29349.8(0.5) + 15000 + 11000 = 40674.5 N
2.2-m
cord
Ladrillos
Cable
55°
Eje
25°
3.0 m-
Transcribed Image Text:2.2-m cord Ladrillos Cable 55° Eje 25° 3.0 m-
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