Escoja la respuesta correcta. Dos personas tratan de mover...

Escoja la respuesta correcta. Dos personas tratan de mover una caja de madera sobre un suelo de madera. Uno empuja la caja con 30.1 N en dirección de positivo x mientras que la otra persona hala la caja con 29.2 N a un ángulo 15.7 grados sobre la horizontal. Calcula la aceleración de la caja si esta se mueve horizontalmente. La masa de la caja es 13.5 Kg. De su respuesta a tres cifras significativas. Los coeficientes de fricción son: ?s =0.5, ?k =0.3. 1.55 m/s^2 2.08 m/s^2 4.39 m/s^2 2.23 m/s^2 0.147 m/s^2 -0.295 m/s^2

Transcript

00:01 One of them says that two people try to move a cajahed a material on a soil of a madeira.

00:07 One, a push the caja with 30 .1 newton's in direction positive.

00:15 So, we'll name this, the force 1 in this direction, the magnitude of the force 1 is equal to 30 .1 newtons.

00:34 While the other person hal a la caja with a force that we're going to call the force f2 and its magnitude is equal to 29 .2 newtons and has an angle of 15 .7 grados we'll let's this angle a lot it would be 15 7 degrees with respect to the horizontal.

01:19 This is this angle here.

01:24 And now we need to calculate the acceleration of the case if this is move horizontally.

01:30 The mass of the mass of the cajah is also to 13 .5 kilograms.

01:40 And we need a response between six figures and the coefficients of friction are done.

01:47 The coefficient of friction static is equal to 0 .5 and the coefficient of fription synethical is equal to 0 .3.

02:00 What we're going to do in this type of exercises is, first, to have all the forces that are applied to this body.

02:12 We have the two forces external.

02:15 Now, we have the forces normal, that is always perpendicular to the surface the peson of this mass that is the mass that's the acceleration of gravity and as we're friction, there's a force that that's upon the movement to the left.

02:41 So what we have to do now is to apply the second law of newton.

02:45 First, for the forces in the direction x.

02:51 So we just that for the right is positive.

02:58 So we have the force f1 plus the force f2, but this is the component x of this force.

03:10 So we have to multiply by the coseno of the angle that has given tita.

03:16 And this is the force of friction...

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