Question

(problem reference m.6)
a boy pulls a 5.0 - kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. the boy pulls on the rope with a force of 10.0 n, and the sled moves with constant velocity.
what is the normal force acting on the sled?
○ 49 n
○ zero n
○ 40 n
○ 66 n
○ 58 n

Explanation:

Step1: Calculate the weight of the sled

The weight of the sled \(W = mg\), where \(m = 5.0\space kg\) and \(g=9.8\space m/s^{2}\). So \(W=5.0\times9.8 = 49\space N\)

Step2: Analyze the vertical component of the pulling force

The pulling force \(F = 10.0\space N\), and the vertical component \(F_y=F\sin\theta\), \(\theta = 60.0^{\circ}\). So \(F_y = 10.0\times\sin60^{\circ}=10.0\times\frac{\sqrt{3}}{2}\approx8.7\space N\)

Step3: Apply the equilibrium condition in the vertical direction

In the vertical direction, \(N + F_y=W\) (since the sled is in vertical equilibrium, \(\sum F_y = 0\)). Then \(N=W - F_y\)
Substitute \(W = 49\space N\) and \(F_y\approx8.7\space N\) into the formula, \(N=49 - 8.7=40.3\approx40\space N\)

Answer:

40 N