What is the minimum downward force on the box in the figure that will keep it from slipping? The coefficients of static and kinetic friction between the box and the floor are 0.38 and 0.28, respectively.
Well, to be precise, the minimum downward force on the box would have to be approximately equal to the weight of the box. However, if you ask me, the box is probably also thinking, "Wow, I really don't want to slip and fall!" So it might add a little extra force, just to be safe. I mean, who wants to be known as the box that couldn't keep its grip, right?
To find the minimum downward force on the box that will keep it from slipping, we need to consider the forces acting on the box.
1. Draw a free-body diagram of the box:
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2. Identify the forces acting on the box:
- Weight (mg) acting vertically downward, where m is the mass of the box and g is the acceleration due to gravity.
- Normal force (N) acting perpendicular to the surface of the floor.
- Friction force (f) acting parallel to the surface of the floor.
3. Since the box is on the verge of slipping, the friction force acting on the box is the maximum static friction force (fs).
4. The maximum static friction force (fs) can be found using the equation:
fs = µs * N,
where µs is the coefficient of static friction.
5. The normal force (N) is equal to the weight (mg) because the box is on a horizontal surface.
6. Substitute the values into the equation:
fs = µs * N
= µs * mg,
7. Substitute the values of the coefficients of static and kinetic friction:
fs = 0.38 * mg.
So, the minimum downward force on the box that will keep it from slipping is 0.38 times the weight (0.38mg).
To find the minimum downward force on the box that will keep it from slipping, we need to consider the forces acting on the box and the conditions for equilibrium.
In this case, we have the force of gravity acting downwards on the box, and the force of friction acting upwards. The maximum static friction force can be calculated using the equation:
F_static = μ_static * N
Where:
- F_static is the maximum static friction force
- μ_static is the coefficient of static friction
- N is the normal force acting perpendicular to the floor
In this case, the normal force N is equal to the weight of the box, which is given by:
N = m * g
Where:
- m is the mass of the box
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the value of N into the equation for the maximum static friction force, we have:
F_static = μ_static * m * g
The minimum downward force to keep the box from slipping is equal to the maximum static friction force. Therefore, the minimum downward force F_downward can be calculated as:
F_downward = μ_static * m * g
Now we can substitute the given values of the coefficients of static and kinetic friction:
μ_static = 0.38
μ_kinetic = 0.28
And solve for the minimum downward force F_downward.
Given: Fs = 0.38, Fk = 0.28, m = ?.
Did you forget the mass?
Mass = 30kg(Assumed).
Wb = m*g = 30kg * 9.8N/kg = 294 N. =
Weight of box.
Fb = 294N @ 0 Deg. = Force of box.
Fp=294*sin(0)=0=Force parallel to floor
Fv = 294*cos(0) = 294 N. = Force perpendicular to floor.
Fn = Fp - u*Fv -u*Fv' = m*a. a = 0.
0 - 0.38*294 - 0.38Fv' = 0
-0.38Fv' = 0.38*294 = 111.72
Fv' = - 294 N.