Question

the box in the diagram above is moving to the right. the tension force is 50 n (x - component = 40 n, y - component = 30 n). if the angle of the tension force (θ) is increased how are friction force and normal force affected? assume the object is moving to the right
a normal force decreases, friction decreases
b normal force decreases, friction increases
c normal force increases, friction decreases
d normal force increases, friction increases

Explanation:

1. Analyze the vertical (y - direction) forces: The normal force \( N \), the weight \( Mg \), and the vertical component of tension \( T_y=T\sin\theta \) (downward). So, \( N + T\sin\theta=Mg \), which can be rewritten as \( N = Mg - T\sin\theta \). When \( \theta \) increases, \( \sin\theta \) increases (for \( 0<\theta < 90^{\circ} \)), so \( N=Mg - T\sin\theta \) decreases.
2. Analyze the horizontal (x - direction) forces: The box is moving to the right, so the friction force \( f \) (opposing motion, to the left) and the horizontal component of tension \( T_x = T\cos\theta \) (to the right). Since the box is moving, we assume kinetic friction, and \( f=\mu N \). We already found that \( N \) decreases as \( \theta \) increases. Since \( \mu \) (coefficient of kinetic friction) is constant, \( f=\mu N \) will also decrease as \( N \) decreases.

Answer:

A. Normal force decreases, friction decreases