say m
normal force m g
Friction force = .5 mg
so
T - .5mg = m a
T = m (1 + .5g) = 5.9 m
The box is being pulled by a roope across a level floor as shown in the diagram. The box is accelerating at 1 m/s2, the coefficient of kinetic friction is 0.5. what is th tension in the rope?
3 answers
sorry, I forgot to inlcude the mass and the angle given. The mass is 10 kg and the angle is 45 degrees
M*g = 10 * 9.8 = 98 N. = Wt. of box.
Fp = 98*sin 0 = 0. = Force parallel with the surface.
Fn = 98*Cos 0 - T*sin45 = 98 - 0.707T = Normal force.
Fk = u*Fn = 0.5(98-0.707T) = 49-0.354T.
T*Cos45-Fp-Fk = M*a
0.707T-0-(49-0.354T) = 10*1,
0.707T+0.354T = 59, T = 55.6 N.
Fp = 98*sin 0 = 0. = Force parallel with the surface.
Fn = 98*Cos 0 - T*sin45 = 98 - 0.707T = Normal force.
Fk = u*Fn = 0.5(98-0.707T) = 49-0.354T.
T*Cos45-Fp-Fk = M*a
0.707T-0-(49-0.354T) = 10*1,
0.707T+0.354T = 59, T = 55.6 N.