A deli owner creates a lunch special display board by taking a uniform board whi

Question

A deli owner creates a lunch special display board by taking a uniform board which has a weight of 127 N, cutting it in half, hinging the halves together with a frictionless hinge, and setting it up as an inverted "V". Determine the minimum coefficient of static friction needed between the board and the ground in order for her to set the display board up with an angle of 30 degree between the two sides. A person is standing on a section of uniform scaffolding as shown in the figure. The section of scaffolding is L = 1.50 m in length, has a m_s = 20.5 kg mass and is supported by three ropes as shown. Determine the magnitude of the tension in each rope when a person with a weight of W_p = 840 N is a distance d = 0.700 m from the left end. magnitude of T_1 vector N magnitude of T_2 vector N magnitude of T_3 vector N

Explanation / Answer

Call the length of each half of the board L.

Draw a clear, large diagram. For an angle 30°between the sides, each board must be 15° to the vertical.

By symmetry, the total weight (127N) acts vertically downwards through the center line. The only other external vertical forces are the normal forces (Fn) at each edge in contact with the ground. Mark all these forces on the diagram.

The normal upwards reaction force at each of the 2 edges is Fn.
Resolving in a vertical direction, 2Fn = 127
Fn = 63.5 N

On the point of slipping, the frictional force at each edge will be Fn = 63.5 where is the coefficient of static friction.

These frictional forces (63.5) both act horizontally inwards. Mark them on the diagram.

Now consider one half of the board in isolation, e.g. the left half.

We need to find torques about the hinge for this half (the internal reaction between the 2 halves at the hinge passes through the hinge so it has zero torque out the hinge).

The 3 forces on the left half, producing a torque about the hinge are the weight of the half board (127/2 = 63.5N), the frictional force,63.5 and the normal force Fn.

From your diagram you should see:

- the torque produced by the weight of the left half board is 63.5*(L/2)sin(15°) (anticlockwise)

- the torque produced by the friction is 63.5*Lcos(15°) (anticlockwise)

- the torque produced by Fn is Fn*Lsin(15°) = 63.5Lsin(15°) (clockwise)

Since the system is in equilibrium:
63.5*(L/2)sin(15°) + 63.5*Lcos(15°) = 63.5Lsin(15°)
(1/2)sin(15°) + cos(15°) = sin(15°)
cos(15°) = (1/2)sin(15°)
= (1/2)tan(15°)
= 0.134

b) First, ignore the person, the weight of the scaffold would be evenly distributed amongst the two ropes. So each rope would then have a tension force of 20.5g/2 = 10.25g N.

Now, using the person as the center of the scaffold. He is 0.7 meter away from one end while 0.8 meters away from the other. I believe it is Kepler's law that I am using here where the the force times distance of one end has to match the force times distance of the other end.

So, to make them equal:
F * 0.7m = (840N - F) * 0.8m
Solving for F:
672N = 1.5F
F = 448 N on the closer end due to the washer and 392 N on the other end.

Adding back the tension force due to the scaffold:
Force on closer end T2= 448+10.25g = 548.55 N
Force on end 0.8 away = 492.55 N

T1*Sin40 = 492.55

T1 = 766.28 N

T1*Cos40 = T3

T3 = 587 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.