Contact forces
Problem:
A 70.0 kg box is pulled across a frictionless surface by a 300 N at an angle of 30° to the horizontal. Complete the following:
Find the acceleration in the x-direction.
Find the normal force.
Calculate the friction required to move the box at a constant speed.
Find the coefficient of friction.
Find the acceleration in the x-direction.
Find the normal force.
Calculate the friction required to move the box at a constant speed.
Find the coefficient of friction.
equations and explanations:
X-DIRECTION ACCELERATION
This solves for the horizontal force on the block represented by x:
cos(30°) = (x/300)
x = 259.8 N
This is the equation F = m*a with F as 259.8 N and a mass of 70.0 kg to solve for acceleration:
-259.8 N = (70.0 kg)*a
a = 3.71 m/s^2NORMAL FORCE
To find the normal force, we use the sigma force equation, which is ΣF = m*a. The sigma force them splits into the force in the y-direction, added to the Normal force, which is the mass multiplied by the acceleration. Each number is then plugged into their representative locations, and simplified. The force in the y-direction is calculated by multiplying the sine of 30° by the force 300 N. Once simplified, the equation comes down to be the simple equation N = 536 N, making the normal force equal to the force of 536 Newtons.
ΣF = m*a
Fy + N—W = m*0
300 N(sin30°) + N — (70.0kg*9.8 m/s^2) = 70.0 kg*0
N = 536 N
FRICTION AT CONSTANT SPEED
This is because, for constant speed, the acceleration would equal to zero because the forces would be balanced, Therefore the friction would equal the horizontal component of the external force. THis would mean that the friction would mean that friction would be equal to 259.8 Newtons of force.
f = Fx
f = (300 N)(cos30°)
f = 259.8 N
FRICTION COEFFICIENT
To solve for this equation is quite easy, because all that is required, is to plug in and simplify. 259.81 N is substituted for friction, while the number 536 replaces N. Once simplified, the coefficient of friction (µ) equals 0.48.
µ*N = f
µ = f/N
µ = 259.81 N/536 N
µ = 0.48
This solves for the horizontal force on the block represented by x:
cos(30°) = (x/300)
x = 259.8 N
This is the equation F = m*a with F as 259.8 N and a mass of 70.0 kg to solve for acceleration:
-259.8 N = (70.0 kg)*a
a = 3.71 m/s^2NORMAL FORCE
To find the normal force, we use the sigma force equation, which is ΣF = m*a. The sigma force them splits into the force in the y-direction, added to the Normal force, which is the mass multiplied by the acceleration. Each number is then plugged into their representative locations, and simplified. The force in the y-direction is calculated by multiplying the sine of 30° by the force 300 N. Once simplified, the equation comes down to be the simple equation N = 536 N, making the normal force equal to the force of 536 Newtons.
ΣF = m*a
Fy + N—W = m*0
300 N(sin30°) + N — (70.0kg*9.8 m/s^2) = 70.0 kg*0
N = 536 N
FRICTION AT CONSTANT SPEED
This is because, for constant speed, the acceleration would equal to zero because the forces would be balanced, Therefore the friction would equal the horizontal component of the external force. THis would mean that the friction would mean that friction would be equal to 259.8 Newtons of force.
f = Fx
f = (300 N)(cos30°)
f = 259.8 N
FRICTION COEFFICIENT
To solve for this equation is quite easy, because all that is required, is to plug in and simplify. 259.81 N is substituted for friction, while the number 536 replaces N. Once simplified, the coefficient of friction (µ) equals 0.48.
µ*N = f
µ = f/N
µ = 259.81 N/536 N
µ = 0.48