Question

A box of mass 27 kg slides down a 14o slope with a coefficient of friction of 0.16.
a)the weight of the box
b) the normal force on the box
c) the frictional force on the box
d) the total force on the box

Answer 1

Answer on Question#37446 - Math - Algebra

A box of mass $27\mathrm{kg}$ slides down a $14{}^{\circ}$ slope with a coefficient of friction of 0.16.

a) the weight of the box

b) the normal force on the box

c) the frictional force on the box

d) the total force on the box

Solution. Let us begin by making a drawing and showing the forces on the box. We also introduce a new coordinate system, so that the $x$ axis coincides with the surface of the slope and the $y$ axis is perpendicular to it (indicated in the drawing below with red arrows):

Here $F_{x}$ is the force in the $x$ direction, $F_{y}$ is the force in the $y$ direction, $F_{g}$ is the force of gravity, $F_{N}$ is the normal force and $F_{f}$ is the force of friction. Finally, $\theta$ is the angle of the slope (in our case, $\theta = 14{}^{\circ}$ ).

Let us now find the forces required in the question.

First, note that the weight of the box is defined as the force on the box due to gravity, and for that we have the standard formula:

$$F _ {g} = m * g,$$

where $g$ is the gravitational acceleration. Since the mass of the box is given ( $m = 27 \, \mathrm{kg}$ ), we take the conventional standard value for $g$ ( $g = 9.80665 \, \mathrm{m/s^2}$ ) and obtain

$$F _ {g} = m * g = 2 7 * 9. 8 0 6 6 5 = 2 6 4. 7 7 9 5 5 \approx 2 6 4. 7 8 (\mathrm {N}).$$

We now find the $x$ and $y$ components of the force:

$$F _ {x} = m * g * \sin \theta , F _ {y} = m * g * \cos \theta .$$

Now note that the normal force $F_{N}$ is always perpendicular to the surface of the slope, and since the box is not moving along the $y$ axis,

$$F _ {N} = F _ {y} = m * g * \cos \theta .$$

Since $\cos 14{}^{\circ} \approx 0.9703$ , we have

$$F _ {N} \approx 2 7 * 9. 8 0 6 6 5 * 0. 9 7 0 3 \approx 2 5 6. 9 1 5 5 9 7 \approx 2 5 6. 9 2 (\mathrm {N}).$$

We can now find the frictional force $F_{f}$ by using the standard formula:

$$F _ {f} = \mu * F _ {N},$$

where $\mu$ is the coefficient of friction (in our case, $\mu = 0.16$ ):

$$F_f \approx 0.16 * 256.915579 \approx 41.106496 \approx 41.11 \text{ (N)}.$$

Finally, we need to find the total force on the box. To do this, notice again that $F_y$ and $F_N$ have the opposite direction and the same value; thus, the forces cancel out each other's action. The total force on the box is then determined by $F_x$ and $F_f$. Let us find $F_x$, remembering the formula above and using the fact that $\sin 14{}^\circ \approx 0.2419$:

$$F_x = m * g * \sin \theta \approx 27 * 9.80665 * 0.2419 \approx 64.055971 \approx 64.06 \text{ (N)}.$$

Now we can find the total force on the box:

$$F = F_x - F_f \approx 64.06 - 41.11 = 22.95 \text{ (N)}.$$

Note that according to our calculations, $F_x > F_f$, and so the total force $F$ is parallel to the surface of the slope and points downward, which is also logical — since the box should be sliding down the slope.

**Answer.**

a) The weight of the box is approximately 264.78 N.

b) The normal force on the box is approximately 256.92 N.

c) The frictional force on the box is approximately 41.11 N.

d) The total force on the box is approximately 22.95 N.