James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
Hi good people of this forum, it is me with another mechanics problem. Please see I drew a diagram to aid my explanation.
A box of mass 3kg lies on a rough plane inclined at 40 degrees to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude X Newtons (the line of action of the force is in the same vertical plane as the line of greatest slope of the inclined plane). The coefficient of friction between the box and the plane is 0.3 and the box is in limiting equilibrium and on the point of moving up the slope. Find X and thus find the normal reaction between the box and the plane.
here is my attempt:
Friction (F) = normal reaction (R) x coefficient of friction =.3R
R = 3gcos40 (the weight of the box) + Xsin40 (the vertical component of the horizontal force)
The box is in static equilibrium so the horizontal component of the horizontal force should be equivalent to the force of friction plus the box's tendency to slide down the slope under its weight:
Xcos40 = F + (3g*sin40)
so Xcos40 = .3(3g*cos40+Xsin40)+(3g*sin40)
rearranging, X = (.9g*cos40+3g*sin40)/(cos40-.3*sin40) = -18, which is very far from the actual answer. What did I do wrong?
A box of mass 3kg lies on a rough plane inclined at 40 degrees to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude X Newtons (the line of action of the force is in the same vertical plane as the line of greatest slope of the inclined plane). The coefficient of friction between the box and the plane is 0.3 and the box is in limiting equilibrium and on the point of moving up the slope. Find X and thus find the normal reaction between the box and the plane.
here is my attempt:
Friction (F) = normal reaction (R) x coefficient of friction =.3R
R = 3gcos40 (the weight of the box) + Xsin40 (the vertical component of the horizontal force)
The box is in static equilibrium so the horizontal component of the horizontal force should be equivalent to the force of friction plus the box's tendency to slide down the slope under its weight:
Xcos40 = F + (3g*sin40)
so Xcos40 = .3(3g*cos40+Xsin40)+(3g*sin40)
rearranging, X = (.9g*cos40+3g*sin40)/(cos40-.3*sin40) = -18, which is very far from the actual answer. What did I do wrong?
