Question

A uniform bar of length 34 m and mass 3 kg is attached to a wall with a hinge that exerts a horizontal force Hx and a vertical force Hy on the bar. The bar is held by a cord that makes a 90◦ angle with respect to the bar and 47 ◦ with respect to the wall.What is the magnitude of the horizontal force Hx on the pivot? The acceleration of gravity is 9.8 m/s 2 . Answer in units of N

Answer 1

Answer:

Magnitude of Hx = 7.33N

Explanation:

From the question,

Length(L) = 34m

Mass(M) = 3kg

Angle with wall (θ) = 47°

acceleration due to gravity (g) = 9.8 m/s²

First of all, let T be the tension in the cord.

Hence, if we take moments about the hinge, we'll obtain :

TL = (MgL/2) cos(θ)

T x 34 = 3 x 9.8 x (34/2)cos47

34T = 3 x 9.8 x 17 x 0.682

34T = 340.86

T = 340.86/34 = 10.025N

So, T = 10.025N

Now, let's resolve horizontal forces to obtain;

Hx = Tsin47= 10.025 x sin 47= 4.89

Hx = 10.025 x 0.7314 = 7.33N

Answer 2

The magnitude of the horizontal force will be "7.33 N".

Given values:

• Length, L = 34 m
• Mass, M = 3 kg
• Angle, $\Theta$ = 47°
• Acceleration due to gravity, g = 9.8 m/s²

By using,

→ $TL = Mg (\frac{L}{2} ) Cos \Theta$

By substituting the values, we get

 $34T=3\times 9.8\times 17\times 0.682$

    $T = \frac{340.86}{34}$

        $= 10.025 \ N$

hence,

The horizontal force will be:

→ $H_x = T Sin 47^{\circ}$

        $=10.025\times 0.7314$

        $= 7.33 \ N$

Thus the above approach is right.

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