Question

an (11.6\text{ kg}) sled is pulled with a (31.2\text{ n}) force at a (46.3^{circ}) angle, across the ground where the coefficient of friction ((mu_k)) is equal to (0.210).

what is the normal force acting upon the sled?

? n

Explanation:

⚡ Using what you learned: newton's laws of motion

Step 1: Identify the forces acting vertically

The sled is on flat ground, so we analyze the forces acting in the vertical direction (\(y\)-axis).

• Gravity pulling downward:

$$ F_g = m \cdot g $$

• Normal force from the ground pushing upward: \(F_N\)
• The vertical component of the pulling force pulling upward:

$$ F_{py} = F_p \cdot \sin(\theta) $$

Step 2: Set up the vertical equilibrium equation

Since the sled does not lift off the ground or sink into it, the net vertical acceleration is zero (\(a_y = 0\)).

$$ \sum F_y = F_N + F_p \cdot \sin(\theta) - m \cdot g = 0 $$

Solve for the normal force \(F_N\):

$$ F_N = m \cdot g - F_p \cdot \sin(\theta) $$

Step 3: Calculate the value

Substitute the given values into the equation:

• \(m = 11.6\text{ kg}\)
• \(g = 9.8\text{ m/s}^2\) (or \(9.81\text{ m/s}^2\))
• \(F_p = 31.2\text{ N}\)
• \(\theta = 46.3^\circ\)

$$ F_N = (11.6 \cdot 9.8) - 31.2 \cdot \sin(46.3^\circ) $$

$$ F_N = 113.68 - 31.2 \cdot 0.72297 $$

$$ F_N = 113.68 - 22.56 $$

$$ F_N \approx 91.12\text{ N} $$

Using \(g = 9.8\text{ m/s}^2\), the normal force is approximately \(91.1\text{ N}\). (If using \(g = 9.81\text{ m/s}^2\), \(F_N = 113.80 - 22.56 = 91.24\text{ N}\)).

Answer:

91.1