A box is being pulled along a rough horizontal surface by a rope which has a tension of 10N at an angle of 60 degrees with the horizontal. If the box moves with constant velocity find the frictional force and the normal reaction force given the weight of the box is 100 N
Take upwards and direction of travel as positive.
Horizontally:
Fa + Ff = 0
10cos(60) N + Ff = 0
Ff = -5 N
Vertically:
Fa + Fg + Fn = 0
10sin(60) N + (-100) N + Fn = 0
Fn = 100 - 5sqrt(3) N
M*g = 100.
M = 100/g = 100/9.8 = 10.2kg = Mass of the box.
Fn = 100 - 10*sin60 = 91.34 N. = Normal force
Fap-Fk = M*a.
10*Cos60-Fk = 10.2*0,
Fk = 10*Cos60 = 5 N. = Force of kinetic friction.
To solve this problem, we can break the tension force into its horizontal and vertical components and then use Newton's second law to find the frictional force and the normal reaction force.
1. Find the horizontal component of the tension force:
The horizontal component of the tension force can be found using the formula:
F_h = F * cos(theta), where F is the tension force (10N) and theta is the angle (60 degrees).
F_h = 10N * cos(60 degrees) = 5N
2. Find the vertical component of the tension force:
The vertical component of the tension force can be found using the formula:
F_v = F * sin(theta), where F is the tension force (10N) and theta is the angle (60 degrees).
F_v = 10N * sin(60 degrees) = 8.66N (approx.)
3. Find the frictional force:
Since the box is moving with constant velocity, the frictional force must be equal in magnitude but opposite in direction to the horizontal component of the tension force.
So, the frictional force is 5N and it acts in the opposite direction to the motion.
4. Find the normal reaction force:
The normal reaction force is the force exerted by the surface on the box and it acts perpendicular to the surface.
Since the box is not accelerating vertically, the net vertical force must be zero.
The weight of the box (100N) is balanced by the vertical component of the tension force (8.66N approx.) and the normal reaction force (N).
So, N = weight - F_v = 100N - 8.66N ≈ 91.34N.
Therefore, the frictional force is 5N, and the normal reaction force is approximately 91.34N.