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Researchers test canine locomotion by measuring the ground reaction force (the vector sum of the normal force and the propulsion force) as dogs push down and backward with both rear legs. In one trial, a dog generated an instantaneous ground reaction force that was 1.7 times the dog's weight and angled 12 ∘ from vertical.What was the magnitude of the dog's horizontal acceleration at that moment?

Sagot :

Let's make a free-body diagram.

As the problem says, the propulsion force is the horizontal component of the ground reaction force, which is the only horizontal component in the system.

[tex]P=F\cdot\cos 78[/tex]

Where F = 1.7mg, let's find F.

[tex]\begin{gathered} P=1.7mg\cdot\cos 78=1.7\cdot m\cdot9.8\cdot\cos 78 \\ P=3.46m \end{gathered}[/tex]

So, the propulsion force is 3.46 times the mass of the dog. Now, let's use Newton's Second Law.

[tex]\begin{gathered} \Sigma F_x=ma_x \\ P=ma_x \\ 3.46m=ma_x \\ a_x=3.46\cdot\frac{m}{s^2} \end{gathered}[/tex]

Therefore, the horizontal acceleration is 3.46 m/s^2.

Now let's use Newton's Second Law for vertical forces.

[tex]\begin{gathered} \Sigma F_y=ma_y \\ N-W=ma_y_{} \\ F\cdot\sin 78-W=ma_y \\ 1.7mg\cdot\sin 78-mg=ma_y \\ mg(1.7\cdot\sin 78-1)=ma_y \\ g(1.7\cdot\sin 78-1)=a_y \\ a_y=9.8(1.7\cdot\sin 78-1) \\ a_y\approx6.5\cdot\frac{m}{s^2} \end{gathered}[/tex]

Therefore, the magnitude of the vertical acceleration is 6.5 m/s^2.

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