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4 years ago

What is the difference between balanced and unbalanced forces

Physics

1 answer:

max2010maxim [7]4 years ago

4 0

Answer:

Explanation:

When the two or more forces acting on a body such tat the net force on the body or the acceleration of the body is zero, then such forces are balanced forces.

For example, in the game of tug of war, both the teams pulls the rope with same force but in opposite directions so that the net acceleration is zero it means a balanced force is acting on the rope.

When tow or more forces acting on a body and there is some net force or acceleration of the body, then the forces are said to be unbalanced.

For example, if we push a box along the plane, the net force is non zero and the forces are unbalanced.

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You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of

Nezavi [6.7K]

Hi there!

For an object on an incline with friction being pulled, the following forces are present.

Force due to Gravity
Force due to Friction
Force due to tension

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
$\Sigma F = F_T + F_f - F_g$

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
$0 = F_T + F_f - F_g$

Now, we can express each force as an equation.

Force due to tension:

Must be solved for.

Force due to gravity:

On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)

This is expressed as:
$F_g = Mgsin\theta$

Force due to friction:

Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:

$N = Mgcos\theta$

$F_f = \mu Mgcos\theta$

Now, plug these expressions into the above equation.

$0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T$

Mg = 245 N (weight). Plug in all values:
$245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}$

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

$N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}$

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