Question

a box having a mass of 50 kg is dragged across a horizontal floor by means of a rope tied on the front of it. the coefficient of friction between the box and the floor is .3. if the angle between the rope and the floor is 30 degrees, what force must be exerted on the rope to move the box with an acceleration of 2.5 m/second squared

Answer 1

Answer: 148.263 N

Explanation:

If we draw a free body diagram of the describes situation we will have the following:

Net force in x:

$-F_{fr}+Fcos\theta=0$ (1)

Where:

$F_{fr}=\mu_{k}N$ is the friction force

$F$ is the force exerted on the rope to move the box

$\theta=30\°$ is the angle

Net force in y:

$N+Fsin\theta-W=0$ (2)

Where:

$N$ is the Normal force

$W=mg$ is the force due gravity on the box (its weight), being $m=50 kg$ the mass of the box and $g=9.8 m/s^{2}$ is the acceleration due gravity

Rewritting (2):

$N+m a sin\theta-mg=0$ (3)

Where $a=2.5 m/s^{2}$

$N=m(g-a sin\theta)$ (4)

$N=50 kg(9.8 m/s^{2}-(2.5 m/s^{2}) sin30\°)$ (5)

$N=428 N$ (6) This is the value of the normal force

Rewritting (1):

$-\mu_{k}N+Fcos\theta=0$ (6)

Finding $F$:

$F=\frac{\mu_{k}N}{cos 30\°}$ (7)

$F=\frac{(0.3)(428 N)}{cos 30\°}$

Finally:

$F=148.263 N$