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4 years ago

Could someone explain how to calculate the frictional force if a side force is placed upon an object?

Physics

1 answer:

tangare [24]4 years ago

5 0

Answer:

Friction force is equal to the coefficient of friction times the normal force.

Explanation:

It does not really matter if the force is side or not. You need to draw a free body diagram, add weight, normal force, friction force and active force. Then see the horizontal and vertical components to calculate what is the normal force.

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Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto

lapo4ka [179]

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

$F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma$

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

$F = ma\\f_s = ma$

That is another equation with one more unknown: fs.

Free-body diagram 2:

$F = (2m)a\\T - f_s = 2ma$

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

$T - ma = 2ma\\T = 3ma$

Plugging this equation into the first equation gives

$T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2$

Following this, we can find the other unknowns:

$T = 3ma = 3m(g/2) = \frac{3mg}{2}$

$f_s = ma = mg/2$

The contact forces are N1 and N2.

$N_1 = mg\\N_2 = 2mg + N1 = 3mg$

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

$y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}$

6 0

4 years ago

What happens to the position of an object as an unbalanced force acts on it? Give an example.

Evgesh-ka [11]

Hi! I'm a few days late since I just saw this question just now, but I'll answer anyway.

Answer:

When an unbalanced force acts on a moving object, the velocity of the object will change. Change in velocity means a change in speed, direction, or both. For example, if you kick a soccer ball and it moves from one place to another, it means that unbalanced forces are acting upon it. The ball moves from one place to another after kicking it so the position changed.

8 0

3 years ago

The impulse given to a body of mass 1.5 kg, is 6.0 kg

yulyashka [42]

Answer:

12J

Explanation:

Given parameters:

Mass = 1.5kg

Impulse = 6kgm/s

let us start first by find the velocity with which this body moves;

Impulse = mass x velocity

Velocity = Impulse / mass = 6/ 1.5 = 4m/s

Initial velocity = 0m/s

Unknown:

Resulting kinetic energy = ?

Solution:

To solve this problem use the formula below:

K.E = $\frac{1}{2}$ m (v - u)²

m is the mass

v is the final velocity

u is the initial velocity

So;

K.E = $\frac{1}{2}$ x 1.5 x (4 - 0)²

K.E = 1.5 x 8 = 12J

6 0

3 years ago

Which of these indicates that a liquid has transferred thermal energy to the air?

Step2247 [10]

The liquid decreases in temperature, and its particles lose kinetic energy.

3 0

3 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago