In a tractor pull competition, a mass m is being dragged across the ground w hor

Question

In a tractor pull competition, a mass m is being dragged across the ground w horizontal acceleration a through a cable (tension T) making an angle theta with respect to the horizontal direction. Meanwhile, there is a kinetic friction coefficient u_k between the mass and the ground. How would you write Newton’s second law in the x direction(along the ground)?

A. Tcos theta -uk(mg-Tsin theta) = 0 B. Tcos theta =ma C. Tcos theta -uk (mg+Tsin theta) = 0 D. Tsin theta = mg E. Tcos theta - uk mg = ma
In a tractor pull competition, a mass m is being dragged across the ground w horizontal acceleration a through a cable (tension T) making an angle theta with respect to the horizontal direction. Meanwhile, there is a kinetic friction coefficient u_k between the mass and the ground. How would you write Newton’s second law in the x direction(along the ground)?

A. Tcos theta -uk(mg-Tsin theta) = 0 B. Tcos theta =ma C. Tcos theta -uk (mg+Tsin theta) = 0 D. Tsin theta = mg E. Tcos theta - uk mg = ma


A. Tcos theta -uk(mg-Tsin theta) = 0 B. Tcos theta =ma C. Tcos theta -uk (mg+Tsin theta) = 0 D. Tsin theta = mg E. Tcos theta - uk mg = ma

Explanation / Answer

E. Tcos theta - uk mg = ma

Net force acting along +x direction, Fnetx = T*cos(theta) - fk

m*a = T*cos(theta) - mue_k*N

m*a = T*cos(theta) - mue_k*m*g <<<<<---------------Answer

In fact N = m*g - T*sin(theta)

so, m*a =  T*cos(theta) - mue_k*(m*g - T*sin(theta)) is the correct answer.

But this is not given in the options.

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