A horse of mass 187 kg pulls a cart of mass 200 kg. The acceleration of gravity is 9.8 m/s2 . What is the largest acceleration the horse can give if the coefficient of static friction between the horse’s hooves and the road is 0.612? Answer in units of m/s2.
net force=mass*a
pulling-friction=mass*a
assuming the cart has no friction..
horsemass*g*.612-0=(187+200)a
solve for a.
To find the largest acceleration the horse can give, we need to consider the forces acting on the system. There are two main forces involved: the force applied by the horse and the force of friction between the horse's hooves and the road.
The force applied by the horse can be calculated using Newton's second law, which states that force (F) is equal to mass (m) times acceleration (a). In this case, the mass of the horse is given as 187 kg. Therefore, the force applied by the horse is:
Force by horse = mass of horse × acceleration by horse
Next, we need to consider the force of friction between the horse's hooves and the road. The maximum static friction force can be calculated using the equation:
Maximum static friction force = coefficient of static friction × normal force
The normal force is the force exerted by the ground on the horse and cart system and is equal to the weight of the system. The weight is given by the formula:
Weight = mass × acceleration due to gravity
For the horse and cart system, the total weight is the sum of the horse's weight and the cart's weight.
Finally, we can set up the equation for the maximum acceleration:
Force by horse - Maximum static friction force = (mass of horse + mass of cart) × acceleration
Now, we can solve the equation to find the maximum acceleration.
First, calculate the weight of the horse:
Weight of horse = mass of horse × acceleration due to gravity
Next, calculate the weight of the cart:
Weight of cart = mass of cart × acceleration due to gravity
The total weight of the system is the sum of the weight of the horse and the weight of the cart.
Finally, calculate the maximum static friction force using the coefficient of static friction and the total weight of the system.
Once you've calculated the maximum static friction force, substitute all the values back into the equation for maximum acceleration:
Force by horse - Maximum static friction force = (mass of horse + mass of cart) × acceleration
Rearrange the equation and solve for acceleration:
Acceleration = (Force by horse - Maximum static friction force) / (mass of horse + mass of cart)
Substitute the values into the equation and solve for acceleration. The answer should be in units of m/s².
To find the largest acceleration the horse can give, we need to consider the force of friction between the horse's hooves and the road. The maximum static friction force can be determined using the equation:
\(f_{\text {static max}} = \mu_{\text {static}} \cdot f_{\text {normal}}\)
Where:
\(f_{\text {static max}}\) is the maximum static friction force,
\(\mu_{\text {static}}\) is the coefficient of static friction, and
\(f_{\text {normal}}\) is the normal force acting on the horse's hooves.
The normal force can be calculated as:
\(f_{\text {normal}} = m_{\text {horse}} \cdot g\)
Where:
\(m_{\text {horse}}\) is the mass of the horse, and
\(g\) is the acceleration due to gravity (9.8 m/s^2).
The maximum static friction force can be given by:
\(f_{\text {static max}} = \mu_{\text {static}} \cdot m_{\text {horse}} \cdot g\)
Applying Newton's second law of motion, we have:
\(f_{\text {static max}} = (m_{\text {horse}} + m_{\text {cart}}) \cdot a_{\text {max}}\)
Where:
\(m_{\text {cart}}\) is the mass of the cart, and
\(a_{\text {max}}\) is the maximum acceleration.
Setting these two equations equal to each other, we can solve for \(a_{\text {max}}\):
\(\mu_{\text {static}} \cdot m_{\text {horse}} \cdot g = (m_{\text {horse}} + m_{\text {cart}}) \cdot a_{\text {max}}\)
Rearranging the equation, we get:
\(a_{\text {max}} = \frac{{\mu_{\text {static}} \cdot m_{\text {horse}} \cdot g}}{{m_{\text {horse}} + m_{\text {cart}}}}\)
Plugging in the given values:
\(a_{\text {max}} = \frac{{0.612 \cdot 187 \, \text {kg} \cdot 9.8 \, \text {m/s}^2}}{{187 \, \text {kg} + 200 \, \text {kg}}}\)
Simplifying the expression:
\(a_{\text {max}} = \frac{{114.8448}}{{387}}\)
Calculating the value:
\(a_{\text {max}} \approx 0.296 \, \text {m/s}^2\)
Therefore, the largest acceleration the horse can give is approximately 0.296 m/s².