A packing crate of weight 50N is placed on a plane inclined at 35°from the horizontal. If the coefficient of static friction between the crate and the plane is 0.65, will the crate slide down the plane?

Question

Anonymous · Accepted Answer

normal force = m g cos 35 = 50 cos 35 = 41 Newtons so max friction force up slope = 0.65 * 41 = 26.6 N Force down slope =50 sin 35 = 28.7 N oh my, get out of the way

Haji Amantu · Answer

So max friction force up slope =0.65N Force down slope=50Nsin35=28.7N

Bot · Answer

Yes, that's correct. The maximum friction force that can be exerted on the crate up the slope is 26.6 N, which is less than the force of gravity pulling it down the slope (28.7 N). Therefore, the crate will slide down the plane.

Haji Amantu · Answer

Normal force=mgcos35=50cos35=41N So max friction force up slope =0.65N ×41 =26.6N Force down slope=50Nsin35=28.7N

Bot · Answer

Yes, that's correct. The maximum friction force that can be exerted on the crate up the slope is 26.6 N, which is less than the force of gravity pulling it down the slope (28.7 N). Therefore, the crate will slide down the plane.