A block with mass m1 = 8.8 kg is on an incline with an angle = 27° with respect

Question

A block with mass m1 = 8.8 kg is on an incline with an angle = 27° with respect to the horizontal. For the first question there is no friction between the incline and the block.

The spring is replaced with a massless rope that pulls horizontally to prevent the block from moving. What is the tension in the rope?

Really struggling with this one. So far all the answers i have seen or tried are wrong. Most people use the equation T= m*g*tan(theta) this is not correct. Any help with formulas would be greatly appreciated. coefficient of static friction is 0.233

Explanation / Answer

1) The four forces whose sum must be zero (to prevent the block from moving) are
(a) the weight of the block (downward vertically)
(b) the frictional force (upward along the incline)
(c) the normal force (upward perpendicular to the incline)
(d) the tension in the rope.
Quantifying these, we have
(a) 8.8 kg times 9.8 m/s^2, direction " -j "
(b) (0.233) times N, direction "i cos 31 + j sin 31"
(c) N, whose magnitude will be found from the vertical balance;
the direction of N is "- i sin 27 + j cos 27"
(d) T, the unknown, direction " i "

The vertical balance requires
(8.8 kg)(9.8 m/s^2) = (0.233) N sin 27 + N cos 27
86.24 newtons = 0.996 N
N = 86.58 newtons

Now T can be found from the horizontal balance:
T = N sin 31 - 0.289 N cos 31
= 24.48 newtons, round to 24.5 newtons

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