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A helicopter flies at a constant altitude towing an airborne 65 kg crate as shown in the diagram. The helicopter and the crate only move in the horizontal direction and have an acceleration of 3.0 m/s21) find the vertical component of the tension in the cable (in Newtons). Ignore the effects of air resistance.2) find the magnitude of the tension in the cable (in Newtons). Ignore the effects of air resistance.3) find the angle(with respect to the horizontal) of the tension in the cable (in degrees). Ignore the effects of air resistance.

A Helicopter Flies At A Constant Altitude Towing An Airborne 65 Kg Crate As Shown In The Diagram The Helicopter And The Crate Only Move In The Horizontal Direc class=

Sagot :

Given data:

* The weight of the crate is 65 kg.

* The acceleration of crate and helicopter is,

[tex]a=3ms^{-2}[/tex]

Solution:

(1). The forces acting on the crate is represented as,

The y-component of tension is balancing the weight of the crate.

Thus, the y-component (vertical) of tension is,

[tex]\begin{gathered} T_y=W \\ T_y=mg \end{gathered}[/tex]

Where m is the mass of crate and g is the acceleration due to gravity,

Substituting the known values,

[tex]\begin{gathered} T_y=65\times9.8 \\ T_y=637\text{ N} \end{gathered}[/tex]

Thus, the vertical component of the tesnion is 637 N.

(2). The x-component of tension in the cable is,

[tex]\begin{gathered} T_x=ma \\ T_x=65\times3 \\ T_x=195\text{ N} \end{gathered}[/tex]

Thus, the tension in the cable is,

[tex]\begin{gathered} T=\sqrt[]{T^2_x+T^2_y} \\ T=\sqrt[]{195^2^{}+637^2} \\ T=666.2\text{ N} \end{gathered}[/tex]

Hence, the tesnion in the cable is 666.2 N.

(3). The angle of tension with the horizontal is,

[tex]\begin{gathered} \tan (\theta)=\frac{T_y}{T_x} \\ \tan (\theta)=\frac{637}{195} \\ \tan (\theta)=3.3 \\ \theta=73.14^{\circ} \end{gathered}[/tex]

Thus, the angle made by the tension with the horizontal is 73.14 degree.

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