Question

the diagram shows a box resting on a wedge which has an angle of inclination of 30°. the coefficient of friction is 0.15 and the vector value of ( f_g ) is 196.20 n. find the mass of the box.
a 10 kg
b 15 kg
c 20 kg
d 40 kg

Explanation:

Step1: Recall forces on incline

For a box on an incline, normal force \( F_N = mg\cos\theta \), friction \( F_f=\mu F_N=\mu mg\cos\theta \), and component of weight down incline \( F_{g\parallel}=mg\sin\theta \). Here, \( F_g = 196.20\ N \) (assuming \( F_g \) is the net force? Wait, maybe \( F_g \) is the force down incline plus friction? Wait, no—wait, the problem says \( F_g = 196.20\ N \). Wait, maybe \( F_g \) is the total force, so \( F_g = mg\sin\theta+\mu mg\cos\theta \)? Wait, no, let's check. Wait, the incline angle \( \theta = 30^\circ \), \( \mu = 0.15 \), \( g = 9.81\ m/s^2 \) (standard). Let's denote mass as \( m \).

Step2: Set up force equation

Assuming \( F_g \) is the force along the incline (maybe the applied force? Wait, no, the diagram: \( F_g \) is red, \( F_N \) normal, \( F_f \) friction. Wait, maybe \( F_g \) is the component of weight plus friction? Wait, no, let's re-express. Wait, the problem says "the vector value of \( F_g \) is 196.20 N". Let's assume that \( F_g = mg\sin\theta+\mu mg\cos\theta \)? Wait, no, maybe \( F_g \) is the net force? Wait, no, let's use \( g = 9.81\ m/s^2 \). Let's solve for \( m \):

Wait, maybe \( F_g \) is the force due to gravity along the incline plus friction? Wait, no, let's think again. Wait, the incline angle is \( 30^\circ \), \( \mu = 0.15 \). Let's compute \( \sin30^\circ = 0.5 \), \( \cos30^\circ=\frac{\sqrt{3}}{2}\approx0.866 \).

Let’s assume that \( F_g = mg\sin\theta + \mu mg\cos\theta \) (if \( F_g \) is the total force opposing motion, but maybe it's the weight component? Wait, no, the weight is \( mg \), so \( mg = 196.20\ N \)? Wait, no, \( g = 9.81 \), so \( m = \frac{196.20}{9.81}=20\ kg \). Wait, that's option C? Wait, no, wait: Wait, maybe \( F_g \) is the component along the incline? No, \( mg\sin30^\circ = m*9.81*0.5 = 4.905m \). Friction is \( \mu mg\cos30^\circ = 0.15*m*9.81*0.866 \approx 0.15*9.81*0.866 m \approx 1.27m \). So total force \( F_g = 4.905m + 1.27m = 6.175m \). But \( F_g = 196.20\ N \), so \( m = 196.20 / 6.175 \approx 31.77 \), no. Wait, maybe I messed up. Wait, maybe \( F_g \) is the weight, so \( mg = 196.20\ N \), so \( m = 196.20 / 9.81 = 20\ kg \). Ah! That makes sense. Because \( 9.81*20 = 196.2\ N \). So maybe \( F_g \) is the weight \( mg \), so \( m = F_g / g = 196.20 / 9.81 = 20\ kg \). So the mass is 20 kg, which is option C? Wait, no, option C is 20 kg? Wait, the options: A.10kg, B.15kg, C.20kg, D.40kg. So 196.20 / 9.81 = 20. So the mass is 20 kg.

Answer:

C. 20 kg