Force a man must exert to pull a sled (8334)

, por F_y_Q

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A man pulls a sled up a ramp using a rope attached to the front, as illustrated in the figure.

The sled has a mass of 80 kg. The kinetic friction coefficient between the sled and the ramp is (\mu_k = 0.70), the angle between the ramp and the horizontal is 25^o, and the angle between the rope and the ramp is 35^o. What force must the man exert to keep the sled moving at a constant speed?

P.-S.

To analyze the forces acting on the sled, decompose them according to the system formed by the direction of movement and a perpendicular axis:


The components of the weight and applied force are:

\left p_x = m\cdot g\cdot sen\ 25 ^o \atop p_y = m\cdot g\cdot cos\ 25^o \right \}

\left F_x = F\cdot cos\ 35 ^o \atop F_y = F\cdot sen\ 35^o \right \}

In the direction perpendicular to the ramp, the sum of the forces must be zero. Consider the "y-component" of the weight to solve for the normal force:

p_y = N + F_y\ \to\ \color[RGB]{2,112,20}{\bf N = m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o}}

In the direction of movement, the sum of the forces must also be zero because the sled moves at a constant speed. This gives:

F_x - p_x- F_R = 0\ \to\ \color[RGB]{2,112,20}{\bm{F_x= p_x - F_R}}

The friction force is the product of the normal force and the coefficient of friction:

F\cdot cos\ 35^o = \mu_c(m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o) + m\cdot g\cdot sen\ 25^o

Substitute the known values into the equations to solve for the force exerted by the man:

0.819F = 497.4 - 0.402F + 331.3\ \to\ 1.221F = 828.7\ \to\ \fbox{\color[RGB]{192,0,0}{\bf F= 678.7\ N}}