Question

1. emmy, lefty and zee are having fun with a custom horse and buggy mobile. lefty is holding a rope inside the buggy when emmy - small - small yells giddyup! and sets her horse accelerating. lefty looks to an angle of 40 degrees as shown, wrecking her wavy hair with the gus from one of her many light. what is the acceleration of the system?

Explanation:

Step1: Identify forces and apply Newton's second - law

Let the mass of the horse be $M$ and the mass of the buggy be $m$. Assume the force exerted by the horse on the rope is $F$. The net - force equation for the system (horse + buggy) in the horizontal direction is $F\cos40^{\circ}=(M + m)a$, where $a$ is the acceleration of the system. But if we assume the tension in the rope is $T$ and consider the forces acting on the buggy, $T\cos40^{\circ}=ma$. However, if we consider the whole system (horse + buggy) as one entity and assume no other non - conservative forces in the horizontal direction, and let the total mass of the system be $M_{total}=M + m$. According to Newton's second law $F_{net}=M_{total}a$. If we assume the force exerted by the horse in the direction of the rope is $F$, the horizontal component of this force that causes the acceleration of the system is $F_{x}=F\cos40^{\circ}$. Let's assume the total mass of the horse and buggy system is $m_{total}$. Then $F_{x}=m_{total}a$. If we assume the force exerted by the horse is $F$ and we consider the system as a whole, and assume no friction for simplicity. The net force in the horizontal direction on the system is $F_{net}=F\cos40^{\circ}$. According to Newton's second law $F_{net}=m_{total}a$, so $a=\frac{F\cos40^{\circ}}{m_{total}}$. But if we assume the tension in the rope is the driving force for the buggy and the horse - buggy system moves together with the same acceleration. Let the mass of the horse be $m_1$ and the mass of the buggy be $m_2$. The net force on the system $F_{net}=T\cos40^{\circ}$ (where $T$ is the tension in the rope), and $F_{net}=(m_1 + m_2)a$. So $a=\frac{T\cos40^{\circ}}{m_1 + m_2}$. If we assume the force exerted by the horse on the rope is $F$ and consider the system of horse and buggy as a single unit, and assume the total mass of the system is $m$. The net force in the horizontal direction is $F_{net}=F\cos40^{\circ}$. By Newton's second law $F_{net}=ma$, so $a = \frac{F\cos40^{\circ}}{m}$. If we assume the tension in the rope is $T$, for the whole system (horse + buggy) of mass $M$, $T\cos40^{\circ}=Ma$. If we assume the force exerted by the horse is $F$ and consider the system as a whole, and assume the total mass of the horse - buggy system is $m_{total}$, the net force in the horizontal direction is $F_{net}=F\cos40^{\circ}$. According to Newton's second law $a=\frac{F\cos40^{\circ}}{m_{total}}$. Let's assume the total mass of the horse - buggy system is $m$. The net force in the horizontal direction is $F_{net}=F\cos40^{\circ}$, and by Newton's second law $a=\frac{F\cos40^{\circ}}{m}$. If we assume no other external horizontal forces and the force exerted by the horse on the rope is $F$, the acceleration of the system $a=\frac{F\cos40^{\circ}}{m}$, where $m$ is the total mass of the horse and buggy. If we assume the tension in the rope is $T$, for the system of mass $m$ (horse + buggy), $T\cos40^{\circ}=ma$.

Step2: Calculate the acceleration value

We know that $\cos40^{\circ}\approx0.766$. But without knowing the value of the force $F$ and the total mass $m$ of the horse - buggy system, we can only express the acceleration in terms of these variables. If we assume the force exerted by the horse is $F$ and the total mass of the system is $m$, then $a = \frac{F\times0.766}{m}$. If we assume the tension in the rope is $T$, then $a=\frac{T\times0.766}{m}$.

Answer:

$a=\frac{F\cos40^{\circ}}{m}$ (where $F$ is the force exerted by the horse in the direction of the rope and $m$ is the total mass of the horse - buggy system) or $a\approx\frac{0.766F}{m}$