A 15000-N crane pivots around a friction-free axle at its base and is supported

Question

A 15000-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25 deg angle with the crane (the figure (Figure 1) ). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane.

A:

When the crane is raised to 55 ? above the horizontal holding an 11000-N pallet of bricks by a 2.2-m very light cord, find the tension in the cable.

Express your answer using two significant figures.

B:

Find the horizontal component of the force that the axle exerts on the crane.

Express your answer using two significant figures.

C:

Find the vertical component of the force that the axle exerts on the crane.

Express your answer using two significant figures.

Explanation / Answer

Question 1:

Find the moments around the axle. Take clockwise as positive. For the system to be in balance the sum of the moments should equal zero.

Take the tension force in the cable to be F. This will have horizontal force Fh (pulling left) and a vertical force Fv (pulling down):

Fh = Fsin60

Fv = Fcos60

(note: the 60 degrees is the cable's angle to the verticle)

Therefore taking the moments around the axle:

[15000 x 7cos55] + [11000 x 16cos55] + [Fcos60 x13cos55]

- [Fsin60 x 13sin55] = 0

F = [15000 x 7cos55] + [11000 x 16cos55] /

13 x ( [sin60 x sin55] - [cos60 x cos55] )

= 29336.34535 N = tension force in cable

Question 2:

Again for the system to be balanced the sum of the horizontal forces must equal zero; take the horizontal component of the force acting on the axle to be Ha:

Ha - Fh = 0 (taking forces left to right to be positive)

Ha = Fh

Fh = Fsin60 = 25406 N

Ha = 25406 N

Question 3:

Sum of vertical forces must equal zero for the system to be in balance; take force on axel to be Va:

Va - 15000 - 11000 - Fv = 0

Fv = Fcos60 = 14668.2 N

Va = 40668.2 N

As a check you can take the moments around another point to see if you are correct. For example you can take the moments around the point the cable joins the crane body like so:

[3cos55 x 11000] + [13cos55 x Va]

- [6cos55 x 15000] - [13sin55 x Ha] = 0

and it still should equal zero; so its correct. Also the 2.2m

cable is just there to throw you. The weight of the bricks only

has a vertical component.

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