"A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?"

Question

"A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?"

I can't get this question, and am getting really frustrated. I know Fg is 12.74N, and should egual F1y. I know the angle between F1 and the horizontal is 62 degrees. But now Ihave no idea what to do. Oh, wait; I know F1x=cos62? So now I'm stuck. Please help?

Answer 1

Fg = (1.3kg)(9.8N/kg) = 12.74N

Fg can be resolved into two foces, T along the extension of the string, and Fp, at right angles to it.
Fp = (12.74N)(cos62)=________N, or
Fp = (12.74N)(sin28) =________N

The vector triangle I hope you drew, shows the weight, Fg resolved into two vectors, Fp, the unbalanced force perpendicular to the string, and T (string tension) which is along the extension of the string and cancelled out by the string resistance.

Answer 2

To find the magnitude of the unbalanced force on the object, we can break it down into two components: the force component acting along the vertical direction (F1y) and the force component acting along the horizontal direction (F1x).

First, let's find the force component along the vertical direction (F1y). We can use the gravitational force formula:

Fg = m * g

where Fg is the gravitational force, m is the mass, and g is the acceleration due to gravity.

Given that the mass (m) of the object is 1.3 kg, we can calculate the gravitational force (Fg) using the equation:

Fg = 1.3 kg * 9.8 m/s^2 = 12.74 N

Since the string makes an angle of 28 degrees with the vertical, and we want to find F1y, we need to find the component of Fg along the vertical direction, which can be obtained using trigonometry.

F1y = Fg * sin(angle)

F1y = 12.74 N * sin(28 degrees) = 5.88 N

Now, let's find the force component along the horizontal direction (F1x). We can use the equation you mentioned, F1x = Fg * cos(angle).

F1x = 12.74 N * cos(28 degrees) = 11.21 N

The magnitude of the unbalanced force on the object is given by the resultant of F1x and F1y. We can use the Pythagorean theorem to find this magnitude:

Unbalanced force = sqrt(F1x^2 + F1y^2)

Unbalanced force = sqrt((11.21 N)^2 + (5.88 N)^2)

Unbalanced force ≈ 12.62 N

Therefore, the magnitude of the unbalanced force on the object is approximately 12.62 N.

"A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?"

Fg = (1.3kg)(9.8N/kg) = 12.74N

down load

6.78

To find the magnitude of the unbalanced force on the object, we can break it down into two components: the force component acting along the vertical direction (F1y) and the force component acting along the horizontal direction (F1x).