a force F pushes on a box of mass m that initially is at rest on a rough, horizontal surface. the force F is directed at an angle theta = 30 degreees above the horizontal. the coefficants of friction between the plane and the box are static= .30 and kinetic = .28. if F=50N and m=15kg, what is the acceleration of the bos and the magnitude of the friction force?
Wb = mg = 15kg * 9.8N/kg = 147 N.
Fb = 147N @ oDeg.
Fp = 147*sin(0) = 0 = Force parallel to surface.
Fv = 147cos(0) -50*sin30 = 122 N. = Force perpendicular to surface.
Fn = Fap*cos30 - Fp - Fk,
Fn = 50*cos30 - 0 - 0.28*122,
Fn = 43.3 - 34.2 = 9.14 N. = Net force.
a = Fn/m = 9.14 / 15 = 0.609 m/s^2.
To find the acceleration of the box and the magnitude of the friction force, we need to break down the given information and apply relevant physics principles.
First, let's consider the forces acting on the box:
1. Force F: This force is applied at an angle of 30 degrees above the horizontal. We need to split this force into its horizontal and vertical components. The horizontal component (Fh) will cause acceleration, while the vertical component (Fv) will not affect the motion, as it is perpendicular to the surface.
2. Weight (mg): This force acts vertically downwards and can be calculated by multiplying the mass (m) of the box by the acceleration due to gravity (g ≈ 9.8 m/s²).
3. Friction force (f): There are two types of friction in play here:
a. Static friction (fs): This force opposes the impending motion of the box while it is at rest.
b. Kinetic friction (fk): Once the box starts moving, the friction force changes to kinetic friction, which opposes the motion.
The maximum static friction (fs max) can be calculated using the formula fs max = μs × N, where μs is the coefficient of static friction and N is the normal force. However, in this case, the box is initially at rest, so the static friction force will be equal to the force applied until it reaches its maximum value. Once the maximum value is reached, the box will start moving.
The kinetic friction (fk) force can be calculated using the formula fk = μk × N, where μk is the coefficient of kinetic friction and N is the normal force. Once the box is in motion, the kinetic friction force will be constant and equal to fk.
To find the normal force (N), we need to consider that it is equal in magnitude to the vertical component of the weight (mg), as the box is on a horizontal surface.
Now, let's calculate the individual components and forces:
1. Vertical Component of F (Fv): Fv = F × sin(θ)
Fv = 50 N × sin(30°)
Fv = 25 N (upwards)
2. Horizontal Component of F (Fh): Fh = F × cos(θ)
Fh = 50 N × cos(30°)
Fh = 43.30 N (to the right)
3. Weight (mg): mg = m × g
mg = 15 kg × 9.8 m/s²
mg = 147 N (downwards)
4. Normal Force (N): N = mg
N = 147 N
5. Static Friction (fs): fs = μs × N
fs = 0.30 × 147 N
fs = 44.10 N (opposing motion until it reaches its maximum)
Now, to find the acceleration (a), we need to compare the forces:
The net horizontal force (Fnet) acting on the box is given by Fnet = Fh - fs.
If Fnet > 0, the box will accelerate, but if Fnet < 0, the box will not move.
Fnet = Fh - fs
= 43.30 N - 44.10 N
= -0.80 N (to the left)
Since Fnet is negative, the box will not move, and therefore, the acceleration (a) will be zero.
Finally, to find the magnitude of the kinetic friction force (fk), we use the formula fk = μk × N. In this case, since the box is not moving, the kinetic friction force will also be zero.
Therefore, the acceleration of the box is 0 m/s², and the magnitude of the friction force is 0 N.