Man pulling a cube on a rough surface

  • Thread starter Thread starter Karol
  • Start date Start date
  • Tags Tags
    Cube Surface
AI Thread Summary
A man with a mass of 70 kg pulls a box at a 10-degree angle, with static and kinetic coefficients of friction of 0.6 and 0.3, respectively. The maximum force he can exert is calculated to be 418.38 N, with a vertical component of 72.65 N affecting the box's weight. The initial calculations suggested the box's maximum mass could be 147.4 kg, but further analysis indicates that the vertical force also increases the normal force on the man. The corrected calculations yield a maximum box mass of approximately 164.87 kg. This highlights the importance of considering all forces acting on the system in physics problems.
Karol
Messages
1,380
Reaction score
22

Homework Statement


A man of mass 70[kg] pules a box at an angle 100. the static coefficient of friction between his legs and the floor is 0.6 and the kinetic coefficient between the box and the floor is 0.3
What's the maximum's box's mass

Homework Equations


Friction: f=μN

The Attempt at a Solution


The horizontal force the man can apply:
$$70\cdot 9.81\cdot 0.6=412.02=F\cdot \cos 10^0\rightarrow F=418.38$$
The vertical component of F raises the box: ##F\cdot \sin 10^0=418.02\cdot\sin 10^0=72.65##
the force needed to pull the box:
$$(mg-F\cdot \sin 10^0)0.3=412.02\rightarrow (mg-72.65)0.3=412.02\rightarrow m=147.4[kg]$$
It should be 164.85
 

Attachments

  • Snap1.jpg
    Snap1.jpg
    4.9 KB · Views: 425
Physics news on Phys.org
The vertical component of F does more than just that !
 
The vertical component also presses down on the man:
$$\left\{ \begin{array}{l} (mg-F\cdot \sin10^0)0.3=(70\cdot g+F\cdot \sin10^0)0.6 \\ F\cdot \cos10^0=(mg-F\cdot \sin10^0)0.3 \end{array} \right. $$
$$\rightarrow F=467.88,\ m=164.87\ \surd$$
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top