Asked by santoki
                A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees. 
If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.
            
        If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.
Answers
                    Answered by
            Henry
            
    m*g = 50 * 9.8 = 490 N. = Wt. of crate =
Normal force (Fn).
a. Fs = u*Fn + u*F*sin30
Fs = 0.75*490 + 0.375F = 367.5 + 0.375F
F*cosA-Fs = m*a
F*cos30-(367.5+0.375F) = m*0 = 0
0.866F-367.5-0.375F = 0
0.866F-0.375F = 367.5
0.491F = 367.5
F = 748.5 N
b. Fs = 0.75*490 - 0.75*F*sin30
Fs = 367.5 - 0.375F
F*cos30-(367.5-0.375F) = m*0 = 0
0.866F-367.5+0.375F = 0
1.241F = 367.5
F = 296.1 N.
    
Normal force (Fn).
a. Fs = u*Fn + u*F*sin30
Fs = 0.75*490 + 0.375F = 367.5 + 0.375F
F*cosA-Fs = m*a
F*cos30-(367.5+0.375F) = m*0 = 0
0.866F-367.5-0.375F = 0
0.866F-0.375F = 367.5
0.491F = 367.5
F = 748.5 N
b. Fs = 0.75*490 - 0.75*F*sin30
Fs = 367.5 - 0.375F
F*cos30-(367.5-0.375F) = m*0 = 0
0.866F-367.5+0.375F = 0
1.241F = 367.5
F = 296.1 N.
                    Answered by
            santoki
            
    thanks Henry!
    
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