What is the mass of the heaviest trunk?

Question

you push downward on a boz at an angle of 25 above the horizontal with a force of 750 N .If the box is on a flat surface and the coefficient of static friction is 0.76. what is the mass of the heaviest trunk you will be able to move

Answer 1

First you must calculate the normal force on the box.



N = N_gravity + N_push.



Now N_gravity = mg, but what is N_push?



Well you are pushing down on the box at an angle 25 above the horizontal. Looking at the vector analysis you will see:

N_push = 750*sin(25)



So the total normal force, N = mg+750*sin(25). Now for you to just be able to move the box you want the horizontal pushing force to equal the force of static friction:



F_push = mu_s *N= mu_s*(mg+750*sin(25))



but F_push = 750*cos(25)



Now plugging this in and solving for m we get



m = 1/g[750*cos(25)/mu_s - 750*sin(25)]



plugging in values we have.



m = 58.9 kg

Answer 2

That's quite a nice problem: Newton set the magical formula for these types of problems centuries ago (F = ma). What you need to do first is to equilibrate the forces. At its break point (the point where te block is jus about to move) the sum of all forces are equal to zero. We can divide these forces in the X and Y component so:

If we define Y as vertical ans X as horizontal



F(y) = 0 = N - F*sin(25) - Fg

(Fg = grav. force, N = normal force, F = 750N)

sin(25) makes the projection of the force on the vertical axis



and in the x component you have:

F(x) = 0 = N*us - F*cos(25)

(us is the 0.76)

cos(25) makes the projection of the force on the horizontal axis



You now have two equations ant two unknowns (N and m) use some math to isolate N in the first formula and to substitute it in the second. You now should have a nice formula with only the mass as an unknown variable in it. Use the same math tools as before to isolate m.



Your answer should be equal to:

m = F*(cos(25) - sin(25)) / (g * us)

(g is the gravitational "constant" 9.807 m/s2)



Hope this helps, the maths are all yours ;)



P.S. The physics behind this problem are quite simple; you can't move the block unless you reach the break point where the force along x is greater or equal to the static maximum friction force (N*us), only then will you be able to have an acceleration greater or equal to 0

Answer 3

Static friction is the force which must be overcome to cause an acceleration in the box. So at the point where the box begins to move, the net forces on the box horizontally are zero. These forces are the force applied (Fcosθ) and the force of friction (-μn, minus because it works against the force applied, n is the normal force). From Newton's 2nd law :



∑F = 0 = Fcosθ - μn

Fcosθ = μn---->equation (1)



The vertical forces are the vertical component of the applied force (-Fsinθ), the normal force, n, and the weight of the box (-mg). NSL again :



∑F = 0 = n – mg – Fsinθ

n = mg + Fsinθ---->equation (2)



Plugging (2) into (1), you get :



Fcosθ = μ(mg + Fsinθ)



Solved for m :



m = [F(cosθ - μsinθ)] / μg

= [750N(cos25° - 0.76sin25°)] / (0.76 x 9.8m/s²)

= 59kg (rounded)



Hope this helps.

Answer 4

The net force in both the x- and y-directions is zero at the limit. The force F has a horizontal component that overcomes the force of friction and a vertical (downward) component that contributes to the normal force.



x: F cos Θ - µ N = 0



y: N - mg - F sin Θ = 0



F cos Θ - µ (mg + F sin Θ) = 0



mg = F (cos Θ - µ sin Θ) / µ