Question

a brick is placed on a ramp that is inclined 24° to the horizontal, as shown. the force c the brick exerts against the ramp has magnitude 13 n. the figure is not drawn to scale. complete the following. do not round any intermediate computations, and round your answers to the nearest hundredth. (a) find the magnitude of the force q required to prevent the brick from sliding down the ramp. n (b) find the weight of the brick, represented by force g. n

Explanation:

Step1: Analyze the force components

We know that the force \(C = 13\space N\). The force \(Q\) (parallel - to - ramp component) and \(C\) (perpendicular - to - ramp component) are related to the weight \(G\) (gravitational force). Using trigonometry, if the angle of the ramp with the horizontal is \(\theta=24^{\circ}\).
The force \(Q\) (to prevent sliding) is given by \(Q = C\tan\theta\), and the weight \(G\) is given by \(G=\frac{C}{\cos\theta}\)

Step2: Calculate the magnitude of force \(Q\)

Substitute \(C = 13\space N\) and \(\theta = 24^{\circ}\) into the formula \(Q = C\tan\theta\).
We know that \(\tan(24^{\circ})\approx0.4452\), so \(Q=13\times\tan(24^{\circ})=13\times0.4452 = 5.79\space N\)

Step3: Calculate the weight \(G\)

Substitute \(C = 13\space N\) and \(\theta = 24^{\circ}\) into the formula \(G=\frac{C}{\cos\theta}\).
We know that \(\cos(24^{\circ})\approx0.9135\), so \(G=\frac{13}{\cos(24^{\circ})}=\frac{13}{0.9135}\approx14.23\space N\)

Answer:

(a) \(5.79\space N\)
(b) \(14.23\space N\)