Question

The new baby hippo at the Cheyenne Mountain Zoo is chilling (not moving) on the ramp that leads to his enclosure. He weighs 157kg. The ramp is at a 10 degree angle.How much force is he experiencing due to gravity?What are the parallel and perpendicular forces? What is the Normal Force he is experiencing? Be sure your reasoning is apparent. What is the Friction Force he is experiencing? Be sure your reasoning is apparent.MUST SHOW WORK

Answer 1

To answer this question we need to draw the free body diagram:

In this case we choose the positive direction as shown in the picture.

Now, from Newton's second law we have:

`\vec{F}=m\vec{a}`

where F is the total force and a is the acceleration. Since the hippo is chilling (not moving) this means that the acceleration is zero, then we would have that the resultant force in this case is:

`\vec{F}=\vec{0}`

Since this is a vector equation we can divide it in two scalar equations, one for each direction; in this case we would have:

`\begin{gathered} W_x-f=0 \\ N-W_y=0 \end{gathered}`

We can answer the first two questions now, The force due to gravity is represented by the weight of the hippo, its magnitude is given by:

`W=mg`

since in this case the mass is 157 kg we have:

`\begin{gathered} W=157\cdot9.8 \\ W=1538.6 \end{gathered}`

Therefore the force due to gravity is 1538.6 N.

Since the wight always points to the center of the earth and in this case we have an inclined plane it can be decompose in two components, one pointing in the x direction and the other pointing in the y direction. From the free body diagram we have that the parallel component would be the x component while the y component will be the perpendicular force. Also from the diagram we notice that the normal force is perpendicular and the friction is parallel; summing up we have:

The x component of the weight and the friction are parallel forces.

The y component of the weight and the normal force ara perpendicular forces.

Now, from the y component of Newton's second law we have that:

`N-W_y=0`

or

`N=W_y`

using the triangle in the diagram we have:

`\begin{gathered} N=W\cos \theta \\ N=1538.6\cos 10 \\ N=1515.23 \end{gathered}`

Therefore the normal force is 1515.23 N

From the x component of Newton's second law we have:

`\begin{gathered} W_x-f=0 \\ \text{ or} \\ f=W_x \end{gathered}`

using the triangle we have:

`\begin{gathered} f=W\sin \theta \\ f=1538.6\sin 10 \\ f=267.18 \end{gathered}`

Therefore the friction force is equal to 267.18 N